soo..Prime factorization of 1332:
13322 * 6662 * 9 * 742 * 3 * 3 * 2 * 372^2 * 3^2 * 37
GCD and LCM of 240 and 840.
240 10 * 245 *
GCD= 2^3 * 3 * 5 = 120
LCM = 2^4 * 3 * 5 * 7 = 1680
If you divide both prime numbers starting with the smallest prime number and continuing to increase the number, you will realize that 19133 isn't prime; but 103 is.
The biggest number that thirteen is divisable by is 480316; since that is nine numbers less than a number divisable by it, the remainder is nine.
and that's about all i can do on this blog.
i'm not sure if i should know how to do this stuff..
or look it up. or maybe i'm just not thinkging properly.
Monday, September 27, 2010
Sunday, September 26, 2010
Connor's blog 4
(Please show your work (briefly))
Question 1:
What is the prime factorization of 1332?
Question 2:
What is the gcd and lcm of 240 and 840?
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
Q1: 1332
2 x 666 ZOMG 666, THE NUMBER OF THE BEAST :O
2 x 333
3 x 111
3 x 37
2^2 x 3^2 x 37
Q2: GCD and LCM 240 and 840
GCD: 840-240=600
600- 240=360
360-240=120
240-120=120
LCM: 2^4 x 3 x 5 x 7 = 1680
Q3: 133 isn't prime; because I'm good like that and I checked it with numberrsss
103 is prime; same as the above reason
Q4: 4803925/2 = 2401962.5
4803925/3 = 1601308.3(repeating)
4803925/5 = 960785.0
4803925/13 = 369532.692
how did I figure them out? I used a calculator of course
Question 1:
What is the prime factorization of 1332?
Question 2:
What is the gcd and lcm of 240 and 840?
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
Q1: 1332
2 x 666 ZOMG 666, THE NUMBER OF THE BEAST :O
2 x 333
3 x 111
3 x 37
2^2 x 3^2 x 37
Q2: GCD and LCM 240 and 840
GCD: 840-240=600
600- 240=360
360-240=120
240-120=120
LCM: 2^4 x 3 x 5 x 7 = 1680
Q3: 133 isn't prime; because I'm good like that and I checked it with numberrsss
103 is prime; same as the above reason
Q4: 4803925/2 = 2401962.5
4803925/3 = 1601308.3(repeating)
4803925/5 = 960785.0
4803925/13 = 369532.692
how did I figure them out? I used a calculator of course
Connor's blog 4
(Please show your work (briefly))
Question 1:
What is the prime factorization of 1332?
Question 2:
What is the gcd and lcm of 240 and 840?
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
Q1: 1332
2 x 666 ZOMG 666, THE NUMBER OF THE BEAST :O
2 x 333
3 x 111
3 x 37
2^2 x 3^2 x 37
Q2: GCD and LCM 240 and 840
GCD: 840-240=600
600- 240=360
360-240=120
240-120=120
LCM: 2^4 x 3 x 5 x 7 = 1680
Q3: 133 isn't prime; because I'm good like that and I checked it with numberrsss
103 is prime; same as the above reason
Q4: 4803925/2 = 2401962.5
4803925/3 = 1601308.3(repeating)
4803925/5 = 960785.0
4803925/13 = 369532.692
how did I figure them out? I used a calculator of course
Question 1:
What is the prime factorization of 1332?
Question 2:
What is the gcd and lcm of 240 and 840?
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
Q1: 1332
2 x 666 ZOMG 666, THE NUMBER OF THE BEAST :O
2 x 333
3 x 111
3 x 37
2^2 x 3^2 x 37
Q2: GCD and LCM 240 and 840
GCD: 840-240=600
600- 240=360
360-240=120
240-120=120
LCM: 2^4 x 3 x 5 x 7 = 1680
Q3: 133 isn't prime; because I'm good like that and I checked it with numberrsss
103 is prime; same as the above reason
Q4: 4803925/2 = 2401962.5
4803925/3 = 1601308.3(repeating)
4803925/5 = 960785.0
4803925/13 = 369532.692
how did I figure them out? I used a calculator of course
Jeffers.!
1. Find the prime factorization of 1332
2 666
2 333
3 111
3 37
2^2 * 3^2 *37
2. GCD and LCM of 240 and 840
2 120
2 60
2 50
2 15
3 5
2^4 * 3*15
3. 840
2 420
2 210
2 105
5 21
3 7
2^3*5*7
GCD=2^3 * 3*5=120
LCM=2^4* 3*5*7= 1680
4.are 133 or 103 prime?
133 is not prime, it is divisible by 7 and 19.
103 is prime b/c it has no other factors beside 1 and itself.
You use a factor tree to solve this out, if you are unable to, then its prime
5. 4803925/2
I solved this by using loooong division.
2=1(odd number)
3=1(number adds to give you 31,one over 30 a multiple of 3)
5=0(goes into number evenly because 5 can go into any number evenly that ends in 0 or 5.)
13=9
2 666
2 333
3 111
3 37
2^2 * 3^2 *37
2. GCD and LCM of 240 and 840
2 120
2 60
2 50
2 15
3 5
2^4 * 3*15
3. 840
2 420
2 210
2 105
5 21
3 7
2^3*5*7
GCD=2^3 * 3*5=120
LCM=2^4* 3*5*7= 1680
4.are 133 or 103 prime?
133 is not prime, it is divisible by 7 and 19.
103 is prime b/c it has no other factors beside 1 and itself.
You use a factor tree to solve this out, if you are unable to, then its prime
5. 4803925/2
I solved this by using loooong division.
2=1(odd number)
3=1(number adds to give you 31,one over 30 a multiple of 3)
5=0(goes into number evenly because 5 can go into any number evenly that ends in 0 or 5.)
13=9
Justin 4
1. find the prime factorization of 1332
3 444
4 111
3 37
2^2x3^2x37
2. find the gcd and lcm of 240 and 840
240
3 80
8 10
2 5
2^4x3x5
840
7 120
5 24
3 8
2^3x3x5x7
GCD=2^3x3x5=120
LCM=2^4x3x5x7=1680
3. are 103 and 133 prime?
133 is not
103 is
divide by prime numbers to find out
4.remainder when 4083925 is diveded by 2, 3, 5, and 13
a. divide by 2 R=1
b. divide by 3 R=1 the digits add up to 31
c. divide by 5 R=0
d. divide by 13 R=9
3 444
4 111
3 37
2^2x3^2x37
2. find the gcd and lcm of 240 and 840
240
3 80
8 10
2 5
2^4x3x5
840
7 120
5 24
3 8
2^3x3x5x7
GCD=2^3x3x5=120
LCM=2^4x3x5x7=1680
3. are 103 and 133 prime?
133 is not
103 is
divide by prime numbers to find out
4.remainder when 4083925 is diveded by 2, 3, 5, and 13
a. divide by 2 R=1
b. divide by 3 R=1 the digits add up to 31
c. divide by 5 R=0
d. divide by 13 R=9
Heather's #4
1.) Prime Factorization of 1332:
1332
2(666)
2(333)
3(111)
3(37)
= 2^3 x 3^3 x 7
2.) GCD (240, 840)
*Prime factorizations:
240:
2(120)
2(60)
2(30)
2(15)
3(5)
=2^4 x 3 x 5
840:
2(420)
2(210)
2(105)
3(35)
5(7)
= 2^3 x 3 x 5 x 7
*To find the GCD, you look at the prime factorization of 240 and 840. You cross out the numbers they do not have in common (so for this one you would rule out 7). Then you multiply the numbers in common that have the least exponent.
*So you would take 2^3 x 3 x 5
GCD = 120
LCM (240, 840)
*For this you still use the prime factorization of both numbers but instead you multiply each number there with the highest exponent (and you include every number present, whether they have it in common or not)
*So you would get 2^4 x 3 x 5 x 7
LCM = 1680
3.) Is 133 prime? NO...Is 103 prime? YES!
*I simply used divisibility rules. And 133 is composite because I found that 7 can go into it evenly, whereas nothing can go into 103.
4.) Remainders when 4803925...
*is divided by 2 -- remainder is 1
*divided by 3 -- remainder is also 1
*divided by 5 -- there is NO remainder. 5 goes in evenly (divisibility rules say that 5 can go into any number that ends in a 5 or zero evenly)
*divided by 13 -- remainder is 9
**I found these out simply by dividing 4803925 by every number given, except 5.
1332
2(666)
2(333)
3(111)
3(37)
= 2^3 x 3^3 x 7
2.) GCD (240, 840)
*Prime factorizations:
240:
2(120)
2(60)
2(30)
2(15)
3(5)
=2^4 x 3 x 5
840:
2(420)
2(210)
2(105)
3(35)
5(7)
= 2^3 x 3 x 5 x 7
*To find the GCD, you look at the prime factorization of 240 and 840. You cross out the numbers they do not have in common (so for this one you would rule out 7). Then you multiply the numbers in common that have the least exponent.
*So you would take 2^3 x 3 x 5
GCD = 120
LCM (240, 840)
*For this you still use the prime factorization of both numbers but instead you multiply each number there with the highest exponent (and you include every number present, whether they have it in common or not)
*So you would get 2^4 x 3 x 5 x 7
LCM = 1680
3.) Is 133 prime? NO...Is 103 prime? YES!
*I simply used divisibility rules. And 133 is composite because I found that 7 can go into it evenly, whereas nothing can go into 103.
4.) Remainders when 4803925...
*is divided by 2 -- remainder is 1
*divided by 3 -- remainder is also 1
*divided by 5 -- there is NO remainder. 5 goes in evenly (divisibility rules say that 5 can go into any number that ends in a 5 or zero evenly)
*divided by 13 -- remainder is 9
**I found these out simply by dividing 4803925 by every number given, except 5.
Prompt 4 from Ryan
Prime factorization of 1332:
1332
2 * 666
2 * 9 * 74
2 * 3 * 3 * 2 * 37
2^2 * 3^2 * 37
GCD and LCM of 240 and 840.
240
10 * 24
5 * 2 * 2 * 12
5 * 2 * 2 * 6 * 2
5 * 2 * 2 * 3 * 2 * 2
2^4 * 3 * 5
840
10 * 84
5 * 2 * 2 * 42
5 * 2 * 2 * 2 * 21
5 * 2 * 2 * 2 * 3 * 7
2^3 * 3 * 5 * 7
GCD = 2^3 * 3 * 5 = 120
LCM = 2^4 * 3 * 5 * 7 = 1680
Are 133 or 103 prime?
You have to divide both by prime numbers starting with 2 and keep going until the number you are dividing by is bigger than the quotient.
133/7 = 19
133 is not prime
103 is prime.
R when 4803925 is divided by 2, 3, 5, and 13.
By 2, it will be R1 because it is an odd number.
By 3, it will be R1 because the number add to give you 31, which is one over a multiple of 3.
By 5, it will be R0 because 5 goes into the number evenly.
By 13, I found the biggest number 13 could go in to less than 4803925, and I got 480316. Since 480316 is 9 less than 4803925 then the remander is 9.
1332
2 * 666
2 * 9 * 74
2 * 3 * 3 * 2 * 37
2^2 * 3^2 * 37
GCD and LCM of 240 and 840.
240
10 * 24
5 * 2 * 2 * 12
5 * 2 * 2 * 6 * 2
5 * 2 * 2 * 3 * 2 * 2
2^4 * 3 * 5
840
10 * 84
5 * 2 * 2 * 42
5 * 2 * 2 * 2 * 21
5 * 2 * 2 * 2 * 3 * 7
2^3 * 3 * 5 * 7
GCD = 2^3 * 3 * 5 = 120
LCM = 2^4 * 3 * 5 * 7 = 1680
Are 133 or 103 prime?
You have to divide both by prime numbers starting with 2 and keep going until the number you are dividing by is bigger than the quotient.
133/7 = 19
133 is not prime
103 is prime.
R when 4803925 is divided by 2, 3, 5, and 13.
By 2, it will be R1 because it is an odd number.
By 3, it will be R1 because the number add to give you 31, which is one over a multiple of 3.
By 5, it will be R0 because 5 goes into the number evenly.
By 13, I found the biggest number 13 could go in to less than 4803925, and I got 480316. Since 480316 is 9 less than 4803925 then the remander is 9.
Feroz - Prompt 4
1. Prime factorization of 1332
1332
2 x 666
2 x 3 x 222
2 x 3 x 2 x 111
2 x 2 x 3 x 3 x 37
2. LCM of 240 and 840
240 = 2^4 x 3 x 5
840 = 2^3 x 3 x 7 x 5
= 120
3. Is 133 and 103 prime? And how do I know?
Well, 140 is divisible by 7. I subtracted 7 from 140 and got 133. So it's composite.
And for 103 I used the Sieve of Eratosthenes. And it's prime.
4. Remainder of 4803925 when divided by 2, 3, 5, and 13.
Any odd number divided by 2 has a remainder of 1. So 2: 1
For 3 I took the last two digits (25) and divided it by 3 ( 3 x 8 = 24). So 3: 1
It ends with a 5, so 5: 0
Unfortunately, I do not know a clever way for 13, so I used long division. 13: 9
1332
2 x 666
2 x 3 x 222
2 x 3 x 2 x 111
2 x 2 x 3 x 3 x 37
2. LCM of 240 and 840
240 = 2^4 x 3 x 5
840 = 2^3 x 3 x 7 x 5
= 120
3. Is 133 and 103 prime? And how do I know?
Well, 140 is divisible by 7. I subtracted 7 from 140 and got 133. So it's composite.
And for 103 I used the Sieve of Eratosthenes. And it's prime.
4. Remainder of 4803925 when divided by 2, 3, 5, and 13.
Any odd number divided by 2 has a remainder of 1. So 2: 1
For 3 I took the last two digits (25) and divided it by 3 ( 3 x 8 = 24). So 3: 1
It ends with a 5, so 5: 0
Unfortunately, I do not know a clever way for 13, so I used long division. 13: 9
Promp #4
Question 1:
What is the prime factorization of 1332?
(2,666)->(2,333)->(3,111)->(3,37)
22*32*37
Question 2:
What is the gcd and lcm of 240 and 840?
240=24*3*5; 840=23*3*5*7
gcd=23*3*5
gcd=120
lcm=24*3*5
lcm=840
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not?
Yes, 133 is prime, so is 103. When each is divided by one of the small primes, there is a remainder of 1, 3, or 13.
Question 4:
What is the remainder when 4803925 is divided by 2? What about 3? 5? and 13? How did you figure each out?
4803925/2 = 2401962 with a remainder of 1.
4803925/3 = 1601908 with a remainder of 1.
4803925/5 = 960785 with no remainder.
4803925/13 = 36958 with a remainder of 1.
What is the prime factorization of 1332?
(2,666)->(2,333)->(3,111)->(3,37)
22*32*37
Question 2:
What is the gcd and lcm of 240 and 840?
240=24*3*5; 840=23*3*5*7
gcd=23*3*5
gcd=120
lcm=24*3*5
lcm=840
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not?
Yes, 133 is prime, so is 103. When each is divided by one of the small primes, there is a remainder of 1, 3, or 13.
Question 4:
What is the remainder when 4803925 is divided by 2? What about 3? 5? and 13? How did you figure each out?
4803925/2 = 2401962 with a remainder of 1.
4803925/3 = 1601908 with a remainder of 1.
4803925/5 = 960785 with no remainder.
4803925/13 = 36958 with a remainder of 1.
Prompt 4
QUESTION 1:
1332
2 x 666
2 x 3 x 222
2 x 3 x 2 x111
2 x 2 x 3 x 3 x 37
QUESTION 2:
240
2 x 120
2 x 6 x 20
2 x 2 x 3 x 5 x 4
2 x 2 x 2 x 3 x 2 x 5
2^4 x 3 x 5
840
2 x 420
2 x 2 x 210
2 x 2 x 2 x 105
2 x 2 x 2 x 5 x 21
2 x 2 x 2 x 5 x 3 x 7
2^3 x 3 x 5 x 7
gcd= 2^3 x 3 x 5 = 120
lcm= 2^4 x 3 x 5 x 7 = 1680
QUESTION 3:
No, its divisible by 7 and 19
Yes, it is only divisible by 1 and itself.
To find then out you are going to figure out what each number is divisble by and if it is divisble by more numbers than 1 and itself then is composite, and if not then it is prime.
QUESTION 4:
I did long division for this.
2 = 1
3 = 1
5 = 0
13 = 9
1332
2 x 666
2 x 3 x 222
2 x 3 x 2 x111
2 x 2 x 3 x 3 x 37
QUESTION 2:
240
2 x 120
2 x 6 x 20
2 x 2 x 3 x 5 x 4
2 x 2 x 2 x 3 x 2 x 5
2^4 x 3 x 5
840
2 x 420
2 x 2 x 210
2 x 2 x 2 x 105
2 x 2 x 2 x 5 x 21
2 x 2 x 2 x 5 x 3 x 7
2^3 x 3 x 5 x 7
gcd= 2^3 x 3 x 5 = 120
lcm= 2^4 x 3 x 5 x 7 = 1680
QUESTION 3:
No, its divisible by 7 and 19
Yes, it is only divisible by 1 and itself.
To find then out you are going to figure out what each number is divisble by and if it is divisble by more numbers than 1 and itself then is composite, and if not then it is prime.
QUESTION 4:
I did long division for this.
2 = 1
3 = 1
5 = 0
13 = 9
Mary's Blog 4!
A)what is the prime factorization of 1332
1332
2x666
2x3x222
2x3x2x111
2x2x3x3x37
B)gcd(240,840)
240
2x120
2x6x20
2x2x3x5x4
2x2x2x3x2x5
2^4x3x5
840
2x420
2x2x210
2x2x2x105
2x2x2x5x21
2x2x2x5x3x7
2^3x3x5x7
gcd=2^3x3x5=120
lcm=2^4x3x5x7=1680
C) is 133 prime?
no, its divisible by 7 and 19
is 103 prime?
yes, it has no factors except 1 and itself.
so find this out, you attempt to make a factor tree, and if you can't, you know that its prime, or you can guess and check with a few numbers
D) 4803925/2
i just used regular old division on some paper for this
2=1
3=1
5=0 because it went evenly
13=9
1332
2x666
2x3x222
2x3x2x111
2x2x3x3x37
B)gcd(240,840)
240
2x120
2x6x20
2x2x3x5x4
2x2x2x3x2x5
2^4x3x5
840
2x420
2x2x210
2x2x2x105
2x2x2x5x21
2x2x2x5x3x7
2^3x3x5x7
gcd=2^3x3x5=120
lcm=2^4x3x5x7=1680
C) is 133 prime?
no, its divisible by 7 and 19
is 103 prime?
yes, it has no factors except 1 and itself.
so find this out, you attempt to make a factor tree, and if you can't, you know that its prime, or you can guess and check with a few numbers
D) 4803925/2
i just used regular old division on some paper for this
2=1
3=1
5=0 because it went evenly
13=9
Mal's Whatever post...
Question 1:
What is the prime factorization of 1332?
1332
2*666
2*2*333
2*2*3*111
2*2*3*3*37
2^2*3^2*37
Question 2:
What is the gcd and lcm of 240 and 840?
Prime factorization of each is:
240=2^4**3*5
840=2^3*3*7*5
so the gcd after taking all of the common numbers between the factorization and the lowest of those and multiplying it together(if that makes any sense) gives you:
gcd(240, 840)=120
now for lcm, you take the highest exponent between the two over every numbers...so this gives you:
lcm(240, 840)=1680
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Okay. so what I did for this thing is use the Sieve of Eran...something. So after writing the first couple of rows of the sieve starting from 101, I went through that process where you start with 2 and so on, crossing out all the multiples of prime numbers. Okay> so...the following I believe to be true:
133-composite...it gets canceled out by 7 in the sieve
103-prime. nothing goes into it.
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
So when I saw this, I was extremely tempted to just plug it in my calculator. That said, I did not.
I then began just dividing using long division (imagine that!). However, I do recognize that I can just find the prime factorization of it and put that over the number I'm dividing by. Again though, I relized that for say 5, I could just use my divisibility rules...5 obviously will go into a number evenly if it ends in what? 0 or 5 right? and this number does...so the remainders are as follows:
2:1
3:1
5:0
13:9
What is the prime factorization of 1332?
1332
2*666
2*2*333
2*2*3*111
2*2*3*3*37
2^2*3^2*37
Question 2:
What is the gcd and lcm of 240 and 840?
Prime factorization of each is:
240=2^4**3*5
840=2^3*3*7*5
so the gcd after taking all of the common numbers between the factorization and the lowest of those and multiplying it together(if that makes any sense) gives you:
gcd(240, 840)=120
now for lcm, you take the highest exponent between the two over every numbers...so this gives you:
lcm(240, 840)=1680
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Okay. so what I did for this thing is use the Sieve of Eran...something. So after writing the first couple of rows of the sieve starting from 101, I went through that process where you start with 2 and so on, crossing out all the multiples of prime numbers. Okay> so...the following I believe to be true:
133-composite...it gets canceled out by 7 in the sieve
103-prime. nothing goes into it.
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
So when I saw this, I was extremely tempted to just plug it in my calculator. That said, I did not.
I then began just dividing using long division (imagine that!). However, I do recognize that I can just find the prime factorization of it and put that over the number I'm dividing by. Again though, I relized that for say 5, I could just use my divisibility rules...5 obviously will go into a number evenly if it ends in what? 0 or 5 right? and this number does...so the remainders are as follows:
2:1
3:1
5:0
13:9
Helen's Blog #4
Question 1:
1332:
12x111
3x4x3x37
3x2x2x3x37
2²x3²x37
Question 2:
gcd(240,840)
240:
24x10
3x8x2x5
3x4x2x2x5
3x2x2x2x2x5
240= 2^4x3x5
840:
84x10
4x21x2x5
2x2x3x7x2x5
840= 2³x3x5x7
GCD = 2³x3x5= 120 LCM= 2^4x3x5x7=1680
Question 3:
133 is not a prime because you can divide it by 19 and 7. 103 is a prime because it is only divisible by itself.
Question 4:
Divide by 2: R=1
Divide by 3: R=1
Divide by 5: R=0
Divide by 9: R=9
I kept dividing until I couldn't divide anymore.
1332:
12x111
3x4x3x37
3x2x2x3x37
2²x3²x37
Question 2:
gcd(240,840)
240:
24x10
3x8x2x5
3x4x2x2x5
3x2x2x2x2x5
240= 2^4x3x5
840:
84x10
4x21x2x5
2x2x3x7x2x5
840= 2³x3x5x7
GCD = 2³x3x5= 120 LCM= 2^4x3x5x7=1680
Question 3:
133 is not a prime because you can divide it by 19 and 7. 103 is a prime because it is only divisible by itself.
Question 4:
Divide by 2: R=1
Divide by 3: R=1
Divide by 5: R=0
Divide by 9: R=9
I kept dividing until I couldn't divide anymore.
Kaitlyn's Blog #4
Question 1:
1332
3 x 444
3 x 148
2 x 74
2 x 37
The prime facorization of 1332 is: 2^2 x 3^2 x 37
Queston 2:
240
3 x 80
2 x 40
2 x 20
2 x 10
2 x 5
240= 2^4 x 3 x 5
840
3 x 280
2 x 140
2 x 70
2 x 35
5 x 7
840= 2^3 x 3 x 5 x 7
gcd (240,840)= 2^3 x 3 x 5= 120
lcm (240,840)= 2^4 x 3 x 5 x 7= 1680
Question 3:
133 is not prime, it is divisible by 7 and 19. 103 is prime because it can't be divided evenly by any number except 1 and itself.
Question 4:
2= 1
3= 1
5= 0
13= 9
I divided each of them on a piece and paper, and the last number that came out that i could not do anything with was the remainder.
1332
3 x 444
3 x 148
2 x 74
2 x 37
The prime facorization of 1332 is: 2^2 x 3^2 x 37
Queston 2:
240
3 x 80
2 x 40
2 x 20
2 x 10
2 x 5
240= 2^4 x 3 x 5
840
3 x 280
2 x 140
2 x 70
2 x 35
5 x 7
840= 2^3 x 3 x 5 x 7
gcd (240,840)= 2^3 x 3 x 5= 120
lcm (240,840)= 2^4 x 3 x 5 x 7= 1680
Question 3:
133 is not prime, it is divisible by 7 and 19. 103 is prime because it can't be divided evenly by any number except 1 and itself.
Question 4:
2= 1
3= 1
5= 0
13= 9
I divided each of them on a piece and paper, and the last number that came out that i could not do anything with was the remainder.
post 4
1- 1332 -2^2 x 3^2 x 37
2-666
2-333
3-111
3-37
2. (240,840) gcd = 120 ..... lcm = 1680
gcd-prime factorization multiply all common numbers with least exponent
lcm-prime factorization multiply all numbers with highest exponent
3.133 isn't prime, it is divisible by 7.(it may divisible by more i just stopped there) i just did guess & check.
103 i think is prime because you check all the numbers up to half way way.. which in this case would be 51.5
4.the remainder is one when divided by two.
the remainder is two when divided by three.
the remainder is zero when divided by five.
the remainder is nine when divided by thirteen.
you find the multiple of the number given before the number you are dividing the given number into... and find the difference
2-666
2-333
3-111
3-37
2. (240,840) gcd = 120 ..... lcm = 1680
gcd-prime factorization multiply all common numbers with least exponent
lcm-prime factorization multiply all numbers with highest exponent
3.133 isn't prime, it is divisible by 7.(it may divisible by more i just stopped there) i just did guess & check.
103 i think is prime because you check all the numbers up to half way way.. which in this case would be 51.5
4.the remainder is one when divided by two.
the remainder is two when divided by three.
the remainder is zero when divided by five.
the remainder is nine when divided by thirteen.
you find the multiple of the number given before the number you are dividing the given number into... and find the difference
Blog #4 Stephen
YAY, FALCONS WON THE GAME!! :) jk.
Anyway. Blog..blog...yes, blog.
Question 1:
What is the prime factorization of 1332?
What is the prime factorization of 1332?
1332: 148 x 9
2x74x9
2x2x37x3x3
2^2 x 3^2 x 37
Question 2:
What is the gcd and lcm of 240 and 840?
What is the gcd and lcm of 240 and 840?
gcd (240, 840)
240: 2x120
2x2x60
2x2x2x30
2x2x2x2x15
2x2x2x2x3x5
240: 2^4 x 3 x 5
840: 3x280
3x2x140
3x2x2x70
3x2x2x2x35
3x2x2x2x5x7
840: 2^3 x 3 x 5 x 7
gcd: 2^3 x 3 x 5 = 120
lcm: 2^4 x 3 x 5 x 7 = 1680
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
133 is not prime because it is divisible by 7 and 19. 103 is prime because its only factors are 1 and 130.
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
by 2 = 1
by 3 = 1
by 5 = 0
by 13 = 9
I used the remainder theorem. Divide regularly, then take the integer part and multiply by the divisor. Then subtract that from the original number, and you get the remainder.
Example: Remainder when 13 is divided by 7.
13/7 = 1.857
1 x 7 = 7
13 - 7 = 6
the remainder is 6
Week Four Blog Prompt
Sorry this one is late, I've had a busy weekend and have been preoccupied.
Okay, this week should be short and relatively simple. There are a total of 9 calculations for you to undergo, none of which are difficult nor lengthy. Brief explanation/work will suffice, but I do want to see something.
(Please show your work (briefly))
Question 1:
What is the prime factorization of 1332?
Question 2:
What is the gcd and lcm of 240 and 840?
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
Okay, this week should be short and relatively simple. There are a total of 9 calculations for you to undergo, none of which are difficult nor lengthy. Brief explanation/work will suffice, but I do want to see something.
(Please show your work (briefly))
Question 1:
What is the prime factorization of 1332?
Question 2:
What is the gcd and lcm of 240 and 840?
Question 3:
Is 133 prime? What about 103? How did you find out that it is/is not (without looking it up.)
Question 4:
What is the remainder when 4803925 is divided by 2? What is it when it is divided by 3? By 5? By 13?
How did you figure each of these out?
Sunday, September 19, 2010
Blog #(insert number here)
blah blah blah
this is really a bad weekend
Question 1:
A) n=1: 2=2
n=2: 2+4=6
n=3: 2+4+6=12
n=4: 2+4+6+8=20
n=5: 2+4+6+8+10=30
n=6: 2+4+6+8+10+12=42
n=7: 2+4+6+8+10+12+14=56
n=8: 2+4+6+8+10+12+14+16=72
n=9: 2+4+6+8+10+12+14+16+18=90
n=10: 2+4+6+8+10+12+14+16+18+20=110
B) well, honestly, I have no clue at all in my mind
C) It doesn't. Logic, in any case, is always extremely flawed and cannot be trusted with power, like math.
Question 2:
A) Huhh? This part is seriously confusing me. I don't understand what you're talking about honestly.
B) Well, since I didn't understand it, I couldn't form a formula....
C) Umm....Ermm.......*music plays*.......I'll have to go with "Herp Derp"....final answer......SURVEY SAYS?!....
this is really a bad weekend
Question 1:
A) n=1: 2=2
n=2: 2+4=6
n=3: 2+4+6=12
n=4: 2+4+6+8=20
n=5: 2+4+6+8+10=30
n=6: 2+4+6+8+10+12=42
n=7: 2+4+6+8+10+12+14=56
n=8: 2+4+6+8+10+12+14+16=72
n=9: 2+4+6+8+10+12+14+16+18=90
n=10: 2+4+6+8+10+12+14+16+18+20=110
B) well, honestly, I have no clue at all in my mind
C) It doesn't. Logic, in any case, is always extremely flawed and cannot be trusted with power, like math.
Question 2:
A) Huhh? This part is seriously confusing me. I don't understand what you're talking about honestly.
B) Well, since I didn't understand it, I couldn't form a formula....
C) Umm....Ermm.......*music plays*.......I'll have to go with "Herp Derp"....final answer......SURVEY SAYS?!....
Feroy's Attempt at Prompt #3
Please don't hurt me if I do this wrong.
Anywho, Question 1:
A. What is the sum of blah blah blah something something.
if n = 2, then s = 2 + 4 = 6
if n = 5, then s = 2 + 4 + 6 + 8 + 10 = 30
So s = 2 + 4 + 6 = 12
B. The formula I used for this was n x (n + 1) So say you have n = 6. You would multiply 6 x 7 to get your sum, which is 42. So instead of adding, like in week 2, you're going to multiply them.
C. It's kinda of like what we did last week, but multiplying instead of adding. And requiring you to change the formula a bit. Yeah...
Question 2:
Pretty much the same thing using odd numbers.
A.
if n=1, s = 1
if n=2, s = 1 + 3 = 4
if n=3, s = 1 + 3 + 5 = 9
if n=4, s = 1 + 3 + 5+ 7 = 16
if n=5, s = 1 + 3 + 5+ 7+9 = 25
if n=6, s = 1 + 3 + 5+ 7+9+11 = 36
if n=7, s = 1 + 3 + 5+ 7+9+11+13 = 49
if n=8, s = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64
B. Don't waste your time adding all of this. Just square n. (n^2)
C. This holds true for all numbers substituted for n because all you're doing is multiplying n by itself, or square it, to get your sum.
Anywho, Question 1:
A. What is the sum of blah blah blah something something.
if n = 2, then s = 2 + 4 = 6
if n = 5, then s = 2 + 4 + 6 + 8 + 10 = 30
So s = 2 + 4 + 6 = 12
B. The formula I used for this was n x (n + 1) So say you have n = 6. You would multiply 6 x 7 to get your sum, which is 42. So instead of adding, like in week 2, you're going to multiply them.
C. It's kinda of like what we did last week, but multiplying instead of adding. And requiring you to change the formula a bit. Yeah...
Question 2:
Pretty much the same thing using odd numbers.
A.
if n=1, s = 1
if n=2, s = 1 + 3 = 4
if n=3, s = 1 + 3 + 5 = 9
if n=4, s = 1 + 3 + 5+ 7 = 16
if n=5, s = 1 + 3 + 5+ 7+9 = 25
if n=6, s = 1 + 3 + 5+ 7+9+11 = 36
if n=7, s = 1 + 3 + 5+ 7+9+11+13 = 49
if n=8, s = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64
B. Don't waste your time adding all of this. Just square n. (n^2)
C. This holds true for all numbers substituted for n because all you're doing is multiplying n by itself, or square it, to get your sum.
Alaina's blog, 19 Sept 2010
Question 1:
A) n=1: 2=2
n=2: 2+4=6
n=3: 2+4+6=12
n=4: 2+4+6+8=20
n=5: 2+4+6+8+10=30
n=6: 2+4+6+8+10+12=42
n=7: 2+4+6+8+10+12+14=56
n=8: 2+4+6+8+10+12+14+16=72
n=9: 2+4+6+8+10+12+14+16+18=90
n=10: 2+4+6+8+10+12+14+16+18+20=110
B) looking back at last week's sums they were, starting at n=1,:
1, 3, 6, 10, 15.. With those numbers, to get the next sum number, all you have to do is add the next "n". meaning that to find the sum number after 6, where n=3, add the next "n" which would be 4. And that would give me the correct sum, 10. for this week, since the numbers are only even numbers, there's a bigger jump from one sum number to the next. First looking at only the sum #'s, starting with 2 you add 4, then you add 4, then 6, then 8, etc. to get the next sum number. you must have to multiply something together in order to get the formula. The formula I came up with to find the sums is 2n. it just sort of gets you where you need to be. Using the formula I had last week, I made a minor adjustment for it to work this time. So my formula for this is 2n(n+1)/2. all you have to do to find the sum is plug in n.
C) I'd say it makes sense because it's the same concept as what we did last week, just with a different category of numbers..So you would have to tweek the formula a little bit in order for it to work properly.
Question 2:
A) n=1: 1=1
n=2: 1+3=4
n=3: 1+3+5=9
n=4: 1+3+5+7=16
n=5: 1+3+5+7+9=25
n=6: 1+3+5+7+9+11=36
n=7: 1+3+5+7+9+11+13=49
n=8: 1+3+5+7+9+11+13+15=64
B) looking at the sums, the formula for this would be n^2 because all the sums are perfect squares.
C) This is true for all n because no matter what n is, you're going to multiply it by itself to get your sum every time. So that will not change anything. And there are really no specifics when it comes to the "category" of the number. for example if you are asked to find the sum when n=137, all you do is multiply 137 by itself and you get 18769.
A) n=1: 2=2
n=2: 2+4=6
n=3: 2+4+6=12
n=4: 2+4+6+8=20
n=5: 2+4+6+8+10=30
n=6: 2+4+6+8+10+12=42
n=7: 2+4+6+8+10+12+14=56
n=8: 2+4+6+8+10+12+14+16=72
n=9: 2+4+6+8+10+12+14+16+18=90
n=10: 2+4+6+8+10+12+14+16+18+20=110
B) looking back at last week's sums they were, starting at n=1,:
1, 3, 6, 10, 15.. With those numbers, to get the next sum number, all you have to do is add the next "n". meaning that to find the sum number after 6, where n=3, add the next "n" which would be 4. And that would give me the correct sum, 10. for this week, since the numbers are only even numbers, there's a bigger jump from one sum number to the next. First looking at only the sum #'s, starting with 2 you add 4, then you add 4, then 6, then 8, etc. to get the next sum number. you must have to multiply something together in order to get the formula. The formula I came up with to find the sums is 2n. it just sort of gets you where you need to be. Using the formula I had last week, I made a minor adjustment for it to work this time. So my formula for this is 2n(n+1)/2. all you have to do to find the sum is plug in n.
C) I'd say it makes sense because it's the same concept as what we did last week, just with a different category of numbers..So you would have to tweek the formula a little bit in order for it to work properly.
Question 2:
A) n=1: 1=1
n=2: 1+3=4
n=3: 1+3+5=9
n=4: 1+3+5+7=16
n=5: 1+3+5+7+9=25
n=6: 1+3+5+7+9+11=36
n=7: 1+3+5+7+9+11+13=49
n=8: 1+3+5+7+9+11+13+15=64
B) looking at the sums, the formula for this would be n^2 because all the sums are perfect squares.
C) This is true for all n because no matter what n is, you're going to multiply it by itself to get your sum every time. So that will not change anything. And there are really no specifics when it comes to the "category" of the number. for example if you are asked to find the sum when n=137, all you do is multiply 137 by itself and you get 18769.
Jeff's Blog
Question 1: a.)S=2+4+6+8=20
b.)it is basically the same concept that we learned last week by adding numbers in a pattern.
c.)Reason why it would make sense is because if you continue to add multiples of 2,you will get the next number in the number sequence.
Question 2: n=0:T=0+3=3
n=1:T=1+3=4
n=2:T=2+3=5
n=3:T=3+3=6
n=4:T=4+3=7
n=5:T=5+3=8
n=6:T=6+3=9
n=7:T=7+3=10
b.) its just like you continue to add one number to the n to get the next sequence
c.) this is true b/c if you continue on with the pattern by adding one to n w/out changing the second number then sum will continue to increase by one.
Ex: n=8:T=8+3=11
n=9:T=9+3=12
n=10:T=10+3=13
b.)it is basically the same concept that we learned last week by adding numbers in a pattern.
c.)Reason why it would make sense is because if you continue to add multiples of 2,you will get the next number in the number sequence.
Question 2: n=0:T=0+3=3
n=1:T=1+3=4
n=2:T=2+3=5
n=3:T=3+3=6
n=4:T=4+3=7
n=5:T=5+3=8
n=6:T=6+3=9
n=7:T=7+3=10
b.) its just like you continue to add one number to the n to get the next sequence
c.) this is true b/c if you continue on with the pattern by adding one to n w/out changing the second number then sum will continue to increase by one.
Ex: n=8:T=8+3=11
n=9:T=9+3=12
n=10:T=10+3=13
IMPORTANT: Clarification and Criticisms
Okay guys, really your first two weeks were fine, but the point of this week was to expand upon what you SHOULD have learned from the first week. As far as I know these are not here for you to submit your responses and then forget the blog ever existed. From what I saw I feel that VERY few of you even bothered to look over the first/second week's comments.
There seems to be a decent amount of confusion, so for that I apologize.
For the first sequence, we're talking about the SUM of the first n EVEN integers (positive ones only that is.)
What this means is that each sum will start with 2, then add 4, then add 6, and so on and so on until you have added a total of n numbers.
The second sequence is the exact same thing except with odd numbers instead of even ones! This means that for some n, you will be adding together n numbers!! Please take note of this and correct your post if you did something else.
Furthermore, I cannot stress this enough, read my previous post concerning week 2!!! It contains the correct formula from week 2 and it IS useful (almost essential) for this week's answer!
For part b in both questions, I do not expect something in terms of the previous term. So you cannot say something like "it's the one before it plus 2." You need to come up with a formula in terms of n. (This should be easy to do for question 1 if you use last week's formula and should be easy to do for question 2 if you just look at the results that you have calculated.)
Finally, for question 2c, please try to think about what is going on with the problem. Saying that a pattern continues indefinitely because you have a hunch is NOT okay within the realm of mathematics. You need to come up with some sort of justification (be it formal or not) other than just "a pattern exists for the first 8 terms... so it must continue forever!"
Hint: Use week 2's question combined with question 1.
Finally, I will say again, if ANYTHING is ambiguous to you or if you are just outright confused, either email me or leave a comment!
There seems to be a decent amount of confusion, so for that I apologize.
For the first sequence, we're talking about the SUM of the first n EVEN integers (positive ones only that is.)
What this means is that each sum will start with 2, then add 4, then add 6, and so on and so on until you have added a total of n numbers.
The second sequence is the exact same thing except with odd numbers instead of even ones! This means that for some n, you will be adding together n numbers!! Please take note of this and correct your post if you did something else.
Furthermore, I cannot stress this enough, read my previous post concerning week 2!!! It contains the correct formula from week 2 and it IS useful (almost essential) for this week's answer!
For part b in both questions, I do not expect something in terms of the previous term. So you cannot say something like "it's the one before it plus 2." You need to come up with a formula in terms of n. (This should be easy to do for question 1 if you use last week's formula and should be easy to do for question 2 if you just look at the results that you have calculated.)
Finally, for question 2c, please try to think about what is going on with the problem. Saying that a pattern continues indefinitely because you have a hunch is NOT okay within the realm of mathematics. You need to come up with some sort of justification (be it formal or not) other than just "a pattern exists for the first 8 terms... so it must continue forever!"
Hint: Use week 2's question combined with question 1.
Finally, I will say again, if ANYTHING is ambiguous to you or if you are just outright confused, either email me or leave a comment!
Post 3
So, I just realized that Jordan's the one doing this blog...I'm must have been out of it when that memo was given out. Anyway, let's get on with the math, shall we? Kay, we shall.
Question 1:
a.) What is the sum of the first n positive, non-zero, even integers?
(1) 2= 2
(2) 2+4= 6
(3) 2+4+6= 12
(4) 2+4+6+8= 20
(5) 2+4+6+8+10= 30
(6) 2+4+6+8+10+12= 42
(7) 2+4+6+8+10+12+14= 56
(8) 2+4+6+8+10+12+14+16= 72
(9) 2+4+6+8+10+12+14+16+18= 90
(10) 2+4+6+8+10+12+14+16+18+20= 110
b.) How does this compare to the sum of the first n integers (from last week?) Here I mean how does it compare numerically, e.g. When n=1, how does this some compare to last week's sum? When n=2, how do they compare? etc. By doing this, one should be able to figure out the generic formula using last week's solution.
Okay. So last week, I did point out that in the case of n=3, the sum was n^2 - n. However....in this case it's n^2 + n, when looking at the above sums, this becomes evident to me...now of course.
c.) Why does this make any logical sense? Be careful before jumping to immediate conclusions!
Well, most of the time I lack in the logic department. But I'll give it a whirl. Ummm...this makes "logical sense" because...okay. I'm at a loss.
Question 2:
Let T be sum of the first n positive, non-zero, odd integers?
Similarly to question one, this just means if n=2, T=1+3=4 and so on.
a.) Write out the sums when n=1, 2, 3, 4, 5, 6, 7, and 8?.
Sums? I like these problems..they start out easy then just progress with the whole thinking thing. gahh!
(1) 1= 1
(2) 1+3= 4
(3) 1+3+5= 9
(4) 1+3+5+7= 16
(5) 1+3+5+7+9= 25
(6) 1+3+5+7+9+11= 36
(7) 1+3+5+7+9+11+13= 49
(8) 1+3+5+7+9+11+13+15= 64
Got it? good.
b.) What is the obvious pattern that arises? What is the sum for any generic number of terms, n? I am expected a formula for this part! This one is not complicated at all assuming your partial sums are correct. If you're staring at your sums and don't know the answer to this one, you most likely calculated them incorrectly!
n^2. does it really need an explanation??
c.) Justify that this holds true for all n. This step IS required and is the main question in this week's blog post. An answer of "I know it's true but don't know why" is NOT acceptable for this one!
Hint: Think about either a geometric argument or an algebraic one using last week's question.
This n^2 thing does hold true for all numbers, because if you think of it "logically", because T(n-1) + n+n-1 ..which basically says T(n-1) + 2n - 1. So THIS basically says, that the previous sum plus the 2n - 1 giving you the next odd number to just add to the previous. I think I'm being kind of redundant, but to be honest, I'm at a loss for words...
Question 1:
a.) What is the sum of the first n positive, non-zero, even integers?
(1) 2= 2
(2) 2+4= 6
(3) 2+4+6= 12
(4) 2+4+6+8= 20
(5) 2+4+6+8+10= 30
(6) 2+4+6+8+10+12= 42
(7) 2+4+6+8+10+12+14= 56
(8) 2+4+6+8+10+12+14+16= 72
(9) 2+4+6+8+10+12+14+16+18= 90
(10) 2+4+6+8+10+12+14+16+18+20= 110
b.) How does this compare to the sum of the first n integers (from last week?) Here I mean how does it compare numerically, e.g. When n=1, how does this some compare to last week's sum? When n=2, how do they compare? etc. By doing this, one should be able to figure out the generic formula using last week's solution.
Okay. So last week, I did point out that in the case of n=3, the sum was n^2 - n. However....in this case it's n^2 + n, when looking at the above sums, this becomes evident to me...now of course.
c.) Why does this make any logical sense? Be careful before jumping to immediate conclusions!
Well, most of the time I lack in the logic department. But I'll give it a whirl. Ummm...this makes "logical sense" because...okay. I'm at a loss.
Question 2:
Let T be sum of the first n positive, non-zero, odd integers?
Similarly to question one, this just means if n=2, T=1+3=4 and so on.
a.) Write out the sums when n=1, 2, 3, 4, 5, 6, 7, and 8?.
Sums? I like these problems..they start out easy then just progress with the whole thinking thing. gahh!
(1) 1= 1
(2) 1+3= 4
(3) 1+3+5= 9
(4) 1+3+5+7= 16
(5) 1+3+5+7+9= 25
(6) 1+3+5+7+9+11= 36
(7) 1+3+5+7+9+11+13= 49
(8) 1+3+5+7+9+11+13+15= 64
Got it? good.
b.) What is the obvious pattern that arises? What is the sum for any generic number of terms, n? I am expected a formula for this part! This one is not complicated at all assuming your partial sums are correct. If you're staring at your sums and don't know the answer to this one, you most likely calculated them incorrectly!
n^2. does it really need an explanation??
c.) Justify that this holds true for all n. This step IS required and is the main question in this week's blog post. An answer of "I know it's true but don't know why" is NOT acceptable for this one!
Hint: Think about either a geometric argument or an algebraic one using last week's question.
This n^2 thing does hold true for all numbers, because if you think of it "logically", because T(n-1) + n+n-1 ..which basically says T(n-1) + 2n - 1. So THIS basically says, that the previous sum plus the 2n - 1 giving you the next odd number to just add to the previous. I think I'm being kind of redundant, but to be honest, I'm at a loss for words...
Justin 3
Question 1
A.)n6=2+4+6+8+10+12=42
B.)the pattern is that you add the previous n, or multiply by the next n... 6x7 = 42 n6=42
C.)it does make sense because you are not changing the numbers that are added
Question 2
A.)n1=1
n2=1+3=4
n3=1+3+5=9
n4=1+3+5+7=16
B.)the pattern is the adding numbers equal n^2
C.)this is true because the odd integers in the series add to n^2 everytime
A.)n6=2+4+6+8+10+12=42
B.)the pattern is that you add the previous n, or multiply by the next n... 6x7 = 42 n6=42
C.)it does make sense because you are not changing the numbers that are added
Question 2
A.)n1=1
n2=1+3=4
n3=1+3+5=9
n4=1+3+5+7=16
B.)the pattern is the adding numbers equal n^2
C.)this is true because the odd integers in the series add to n^2 everytime
Post 3
1)
a) What is the sum of the first n positive, non-zero, even integers?
S = (n)(n+1)
b) The sum if only doing even integers will be double than if doing all integers.
c) It makes sense because of the fact that every number you're adding for even integers is exactly double than for all integers.
i.e.
n=6
all: S = 1+2+3+4+5+6 = 21
even: S = 2+4+6+8+10+12 = 42
1*2 = 2, 2*2 = 4, 3*2 = 6, etc.
2)
a)
1. 1=1
2. 1+3 = 4
3. 1+3+5 = 9
4. 1+3+5+7 = 16
5. 1+3+5+7+9 = 25
6. 1+3+5+7+9+11 = 36
7. 1+3+5+7+9+11+13 = 49
8. 1+3+5+7+9+11+13+15 = 64
b)the answer will be S = n^2
c) Another pattern I found that may also be the proof is:
in all n's the first and last number added in the series combine to give you 2n. And then combined with n/2 also gives you n^2.
a) What is the sum of the first n positive, non-zero, even integers?
S = (n)(n+1)
b) The sum if only doing even integers will be double than if doing all integers.
c) It makes sense because of the fact that every number you're adding for even integers is exactly double than for all integers.
i.e.
n=6
all: S = 1+2+3+4+5+6 = 21
even: S = 2+4+6+8+10+12 = 42
1*2 = 2, 2*2 = 4, 3*2 = 6, etc.
2)
a)
1. 1=1
2. 1+3 = 4
3. 1+3+5 = 9
4. 1+3+5+7 = 16
5. 1+3+5+7+9 = 25
6. 1+3+5+7+9+11 = 36
7. 1+3+5+7+9+11+13 = 49
8. 1+3+5+7+9+11+13+15 = 64
b)the answer will be S = n^2
c) Another pattern I found that may also be the proof is:
in all n's the first and last number added in the series combine to give you 2n. And then combined with n/2 also gives you n^2.
Stephen's blog 3?
Question 1:
a) n=3, s=2+4+6=12
n=4, s=2+4+6+8=20
n=5, s=2+4+6+8+10=30
n=6, s=2+4+6+8+10+12=42
and so on...
b) for last week, all that basically happens is you add the previous n. As n gets larger, the value that comes out is the previous value added to n. The pattern is very uniform and easy to figure out, to me at least.
c) It makes perfect logical sense because when you add n or do anything with it basically, you're using the same numbers, therefore getting similar numbers as outputs.
Question 2:
a) n=1, s=1=1
n=2, s=1+3=4
n=3, s=1+3+5=9
n=4, s=1+3+5+7=16
n=5, s=1+3+5+7+9=25
n=6, s=1+3+5+7+9+11=36
n=7, s=1+3+5+7+9+11+13=49
n=8, s=1+3+5+7+9+11+13+15=64
n=9, s=1+3+5+7+9+11+13+15+17=81
b) The obvious pattern that arises is that the sum of n is n^2, which is an obvious formula.
c) This holds true for n because every time you add these odd integers of n itself, you will get n^2. All of these numbers are related. No matter what you plug in, you will always get that number multiplied by itself.
question 1-
a. 5n
b.n x 1 from last week. soooo.. and whatever n is, then it is divisible by that number.
c.it makes logical sense because typically all the formulas are very similar and they all have to just be changed a little to work. so it becomes logical in my opinion.
question 2-
a.
n = 1 - 1 = 1
n = 2 - 1+3 = 4
n = 3 - 1+3+5 = 9
n = 4 - 1+3+5+7 = 16
n = 5 - 1+3+5+7+9 = 25
n = 6 - 1+3+5+7+9+11 = 36
n = 7 - 1+3+5+7+9+11+13 = 49
n = 8 - 1+3+5+7+9+11+13+15 = 64
b. the difference between each number is odd. starting at 3, then 5 then 7.. now do i know how to generate a formula for that? no... oh & multiply the number by itself and you get that number also
c. this is true for all n terms because no matter what the number is if you are asked to find the even sum ... you just multiply the number times itself and get your answer.
a. 5n
b.n x 1 from last week. soooo.. and whatever n is, then it is divisible by that number.
c.it makes logical sense because typically all the formulas are very similar and they all have to just be changed a little to work. so it becomes logical in my opinion.
question 2-
a.
n = 1 - 1 = 1
n = 2 - 1+3 = 4
n = 3 - 1+3+5 = 9
n = 4 - 1+3+5+7 = 16
n = 5 - 1+3+5+7+9 = 25
n = 6 - 1+3+5+7+9+11 = 36
n = 7 - 1+3+5+7+9+11+13 = 49
n = 8 - 1+3+5+7+9+11+13+15 = 64
b. the difference between each number is odd. starting at 3, then 5 then 7.. now do i know how to generate a formula for that? no... oh & multiply the number by itself and you get that number also
c. this is true for all n terms because no matter what the number is if you are asked to find the even sum ... you just multiply the number times itself and get your answer.
Helen's Blog #3
Question #1:
A) S= 2+4+6=12
B) It's the same concept that we did last week by adding numbers in a pattern.
C) It makes logical sense because if you keep adding multiples of 2 you will get the next number in the sequence.
Question #2:
A)n=1: T=1+3=4
n=2: T=2+3=5
n=3: T=3+3=6
n=4: T=4+3=7
n=5: T=5+3=8
n=6: T=6+3=9
n=7: T=7+3=10
n=8: T=8+3=11
B)It's like you just add one to the n to get the next sequence.
C)This is true for n because if you keep going on with the pattern, adding one to n without changing the second number then the sum will just keep increasing by one.
Ex: n=9: T=9+3=12
n=10: T=10+3=13
n=11: T=11+3=14
A) S= 2+4+6=12
B) It's the same concept that we did last week by adding numbers in a pattern.
C) It makes logical sense because if you keep adding multiples of 2 you will get the next number in the sequence.
Question #2:
A)n=1: T=1+3=4
n=2: T=2+3=5
n=3: T=3+3=6
n=4: T=4+3=7
n=5: T=5+3=8
n=6: T=6+3=9
n=7: T=7+3=10
n=8: T=8+3=11
B)It's like you just add one to the n to get the next sequence.
C)This is true for n because if you keep going on with the pattern, adding one to n without changing the second number then the sum will just keep increasing by one.
Ex: n=9: T=9+3=12
n=10: T=10+3=13
n=11: T=11+3=14
Kaitlyn's Post #3
Question 1:
A) n=2 s=2+4=6 / n=5 s=2+4+6+8+10=30
s=2+4+6=12
B) n=1=1 n=2=3 n=3=6 n=4=10
n=1=2 n=2=6 n=3=12 n=4=20 (if you take the sum of the first n integars, and then you take the sum of the first n even integars, you will notice that they double)
C) It makes sense because when you are adding the even numbers, you actually are doubling the sum. All the even numbers are doubled and can be divided by two so this is what happens when you do this.
Question 2:
A) n=1 n=1
n=2 n=1+3=4
n=3 n=1+3+5=9
n=4 n=1+3+5+7=16
n=5 n=1+3+5+7+9=25
n=6 n=1+3+5+7+9+11=36
n=7 n=1+3+5+7+9+11+13=49
n=8 n=1+3+5+7+9+11+13+15=63
B) If you didn't want to do all this adding, all you would have to do it square your number. Each number is the square root of the sums.
C) This holds true for all n because if you kept going with this, you will see that it will continue to square out througout all numbers. Numbers are made to work out the way they do.
A) n=2 s=2+4=6 / n=5 s=2+4+6+8+10=30
s=2+4+6=12
B) n=1=1 n=2=3 n=3=6 n=4=10
n=1=2 n=2=6 n=3=12 n=4=20 (if you take the sum of the first n integars, and then you take the sum of the first n even integars, you will notice that they double)
C) It makes sense because when you are adding the even numbers, you actually are doubling the sum. All the even numbers are doubled and can be divided by two so this is what happens when you do this.
Question 2:
A) n=1 n=1
n=2 n=1+3=4
n=3 n=1+3+5=9
n=4 n=1+3+5+7=16
n=5 n=1+3+5+7+9=25
n=6 n=1+3+5+7+9+11=36
n=7 n=1+3+5+7+9+11+13=49
n=8 n=1+3+5+7+9+11+13+15=63
B) If you didn't want to do all this adding, all you would have to do it square your number. Each number is the square root of the sums.
C) This holds true for all n because if you kept going with this, you will see that it will continue to square out througout all numbers. Numbers are made to work out the way they do.
Heather's #3
Question 1:
A) n=1: 2=2
n=2: 2+4=6
n=3: 2+4+6=12
n=4: 2+4+6+8=20
n=5: 2+4+6+8+10=30
n=6: 2+4+6+8+10+12=42
n=7: 2+4+6+8+10+12+14=56
n=8: 2+4+6+8+10+12+14+16=72
n=9: 2+4+6+8+10+12+14+16+18=90
n=10: 2+4+6+8+10+12+14+16+18+20=110
B) Okay, so looking back at last week's sums I saw they were, starting at n=1,:
1, 3, 6, 10, 15..With those numbers, I noticed that to get the next sum number, all you have to do is add the next "n"..meaning that if I wanted to find the sum number after 6, where n=3, all I would have to do is add the next "n" which would be 4. And that would give me the correct sum, 10. Now for this week, since the numbers are only even numbers, there's a bigger jump from one sum number to the next. First looking at only the sum #'s, I saw that starting with 2 you add 4, then you add 4, then 6, then 8, etc. to get the next sum number..Soooo, I figured you must have to multiply something together in order to get the formula. The simple formula I came up with to find the sums is 2n. now I say "simple" because that isn't the actual formula, it just sort of gets you where you need to be. Using the formula I had last week, I made a minor adjustment for it to work this time. So my formula for this is 2n(n+1)/2 ..where all you have to do to find the sum is plug in n.
C) Well I'd say it makes logical sense because it's basically the same concept as what we did last week, just with a different category of numbers..So therefore, you would have to tweek the formula a little bit in order for it to work properly.
Question 2:
A) n=1: 1=1
n=2: 1+3=4
n=3: 1+3+5=9
n=4: 1+3+5+7=16
n=5: 1+3+5+7+9=25
n=6: 1+3+5+7+9+11=36
n=7: 1+3+5+7+9+11+13=49
n=8: 1+3+5+7+9+11+13+15=64
B) Obviously looking at the sums, the formula for this would simply be n^2 because all the sums are in fact, perfect squares.
C) This holds true for all n because no matter what n is, you're going to multiply it by itself to get your sum every time. So therefore that will not change anything. And there are really no specifics when it comes to the "category" of the number..(as in if it's even, odd, a palindrome, etc.) So for example if you are asked to find the sum when n=137, all you do is multiply 137 by itself and you get 18769.
A) n=1: 2=2
n=2: 2+4=6
n=3: 2+4+6=12
n=4: 2+4+6+8=20
n=5: 2+4+6+8+10=30
n=6: 2+4+6+8+10+12=42
n=7: 2+4+6+8+10+12+14=56
n=8: 2+4+6+8+10+12+14+16=72
n=9: 2+4+6+8+10+12+14+16+18=90
n=10: 2+4+6+8+10+12+14+16+18+20=110
B) Okay, so looking back at last week's sums I saw they were, starting at n=1,:
1, 3, 6, 10, 15..With those numbers, I noticed that to get the next sum number, all you have to do is add the next "n"..meaning that if I wanted to find the sum number after 6, where n=3, all I would have to do is add the next "n" which would be 4. And that would give me the correct sum, 10. Now for this week, since the numbers are only even numbers, there's a bigger jump from one sum number to the next. First looking at only the sum #'s, I saw that starting with 2 you add 4, then you add 4, then 6, then 8, etc. to get the next sum number..Soooo, I figured you must have to multiply something together in order to get the formula. The simple formula I came up with to find the sums is 2n. now I say "simple" because that isn't the actual formula, it just sort of gets you where you need to be. Using the formula I had last week, I made a minor adjustment for it to work this time. So my formula for this is 2n(n+1)/2 ..where all you have to do to find the sum is plug in n.
C) Well I'd say it makes logical sense because it's basically the same concept as what we did last week, just with a different category of numbers..So therefore, you would have to tweek the formula a little bit in order for it to work properly.
Question 2:
A) n=1: 1=1
n=2: 1+3=4
n=3: 1+3+5=9
n=4: 1+3+5+7=16
n=5: 1+3+5+7+9=25
n=6: 1+3+5+7+9+11=36
n=7: 1+3+5+7+9+11+13=49
n=8: 1+3+5+7+9+11+13+15=64
B) Obviously looking at the sums, the formula for this would simply be n^2 because all the sums are in fact, perfect squares.
C) This holds true for all n because no matter what n is, you're going to multiply it by itself to get your sum every time. So therefore that will not change anything. And there are really no specifics when it comes to the "category" of the number..(as in if it's even, odd, a palindrome, etc.) So for example if you are asked to find the sum when n=137, all you do is multiply 137 by itself and you get 18769.
Post #3
Question Number 1:
A. n=2 s=2+4=6 s=2+4+6+8+10=30 answer: s=2+4+6=12
B. Well all of the sums are related because when n=1 it is always going to be disivible by one and it will remain the same number because nx1=n, when applied to n=2 then that must mean everything is divisible by 2 and a multiple of 2 basically just doubling the number.
C. It makes logic sense because when you have a number and you multiply OR divide by one it is always going to give you that same number. Also, same thing applies with the number two if you multiply or divide the number by 2 it has to be an even number and the next number in the sequence could be found by just doubling the number.
Question Number 2:
A.
T=1+3=4
T=2+3=5
T=3+3=6
T=4+3=7
T=5+3=8
T=6+3=9
T=7+3=10
T=8+3=11
B. You are going to basically just add one more to the next number of the equation or series, which ever one you prefer to call it.
C. This holds true for n because if you were to add on to the problem and keep going 9+3=12 which proves that adding one to the first number but not changing the second number will only increase the number by one.
A. n=2 s=2+4=6 s=2+4+6+8+10=30 answer: s=2+4+6=12
B. Well all of the sums are related because when n=1 it is always going to be disivible by one and it will remain the same number because nx1=n, when applied to n=2 then that must mean everything is divisible by 2 and a multiple of 2 basically just doubling the number.
C. It makes logic sense because when you have a number and you multiply OR divide by one it is always going to give you that same number. Also, same thing applies with the number two if you multiply or divide the number by 2 it has to be an even number and the next number in the sequence could be found by just doubling the number.
Question Number 2:
A.
T=1+3=4
T=2+3=5
T=3+3=6
T=4+3=7
T=5+3=8
T=6+3=9
T=7+3=10
T=8+3=11
B. You are going to basically just add one more to the next number of the equation or series, which ever one you prefer to call it.
C. This holds true for n because if you were to add on to the problem and keep going 9+3=12 which proves that adding one to the first number but not changing the second number will only increase the number by one.
Week 3 Post
Question 1:
A.) S = 2+4+6 = 12
B.) It is basically the same concept, you are adding things in a pattern.
C.) It makes logical sense because when you add something you'll get an answer relatively involving some qualities of the number.
Question 2:
A.) n=1, 1+2+3 = 6; n=2, 2+4+6 = 12; n=3, 3+6+9 = 18; n=4, 4+8 + 12 = 24; n=5, 5+10 = 15
B.) a divisor of n
C.) This does hold true for all n terms because that is the cool thing about numbers, they all have some relation to each other.
A.) S = 2+4+6 = 12
B.) It is basically the same concept, you are adding things in a pattern.
C.) It makes logical sense because when you add something you'll get an answer relatively involving some qualities of the number.
Question 2:
A.) n=1, 1+2+3 = 6; n=2, 2+4+6 = 12; n=3, 3+6+9 = 18; n=4, 4+8 + 12 = 24; n=5, 5+10 = 15
B.) a divisor of n
C.) This does hold true for all n terms because that is the cool thing about numbers, they all have some relation to each other.
Friday, September 17, 2010
Week Three Blog Prompt
*A few edits are given in red.
So it's that time again. I'll cut to the chase, I'm not entirely sure what you guys are working on at the moment in the class, so I am just going to pick another problem that is particularly interesting to me. This one is very similar to last week's.
If you'll notice, I've posted some comments about blog #2. you might want to look over them as they will be a great help for this week's.
Both questions are required.
Question 1:
a.) What is the sum of the first n positive, non-zero, even integers?
That is to say if n=2, S=2+4=6. If n=5, S=2+4+6+8+10=30, what is the sum (S) for a generic n?
b.) How does this compare to the sum of the first n integers (from last week?) Here I mean how does it compare numerically, e.g. When n=1, how does this some compare to last week's sum? When n=2, how do they compare? etc. By doing this, one should be able to figure out the generic formula using last week's solution.
c.) Why does this make any logical sense? Be careful before jumping to immediate conclusions!
Question 2:
Let T be sum of the first n positive, non-zero, odd integers?
Similarly to question one, this just means if n=2, T=1+3=4 and so on.
a.) Write out the sums when n=1, 2, 3, 4, 5, 6, 7, and 8?.
b.) What is the obvious pattern that arises? What is the sum for any generic number of terms, n? I am expected a formula for this part! This one is not complicated at all assuming your partial sums are correct. If you're staring at your sums and don't know the answer to this one, you most likely calculated them incorrectly!
c.) Justify that this holds true for all n. This step IS required and is the main question in this week's blog post. An answer of "I know it's true but don't know why" is NOT acceptable for this one!
Hint: Think about either a geometric argument or an algebraic one using last week's question.
If you don't know what to do here, take my advice and read my comments for the previous weeks. In particular, look over last week's section! It will be helpful!
Lastly, remember that if you have any questions or concerns, you can email me at jmartin5@tulane.edu. It is very likely that I can get back to you within an hour or two, so if you are confused or concerned about anything, LET ME KNOW!
Good luck!
So it's that time again. I'll cut to the chase, I'm not entirely sure what you guys are working on at the moment in the class, so I am just going to pick another problem that is particularly interesting to me. This one is very similar to last week's.
If you'll notice, I've posted some comments about blog #2. you might want to look over them as they will be a great help for this week's.
Both questions are required.
Question 1:
a.) What is the sum of the first n positive, non-zero, even integers?
That is to say if n=2, S=2+4=6. If n=5, S=2+4+6+8+10=30, what is the sum (S) for a generic n?
b.) How does this compare to the sum of the first n integers (from last week?) Here I mean how does it compare numerically, e.g. When n=1, how does this some compare to last week's sum? When n=2, how do they compare? etc. By doing this, one should be able to figure out the generic formula using last week's solution.
c.) Why does this make any logical sense? Be careful before jumping to immediate conclusions!
Question 2:
Let T be sum of the first n positive, non-zero, odd integers?
Similarly to question one, this just means if n=2, T=1+3=4 and so on.
a.) Write out the sums when n=1, 2, 3, 4, 5, 6, 7, and 8?.
b.) What is the obvious pattern that arises? What is the sum for any generic number of terms, n? I am expected a formula for this part! This one is not complicated at all assuming your partial sums are correct. If you're staring at your sums and don't know the answer to this one, you most likely calculated them incorrectly!
c.) Justify that this holds true for all n. This step IS required and is the main question in this week's blog post. An answer of "I know it's true but don't know why" is NOT acceptable for this one!
Hint: Think about either a geometric argument or an algebraic one using last week's question.
If you don't know what to do here, take my advice and read my comments for the previous weeks. In particular, look over last week's section! It will be helpful!
Lastly, remember that if you have any questions or concerns, you can email me at jmartin5@tulane.edu. It is very likely that I can get back to you within an hour or two, so if you are confused or concerned about anything, LET ME KNOW!
Good luck!
Sunday, September 12, 2010
Responses to Blog Prompts #1 and #2
Okay everyone, so this is going to contain a few brief comments concerning the first two prompts.
First things first, most everyone did fine with their responses (as like I mentioned, I was only asking for your points of view for these two. Do not expect these to stay QUITE as open ended as the year progresses.) I would like to note that while short answers ARE acceptable, I prefer to see the brief explanation to full paragraph length answers. This really isn't a lot of work to do once a week (it should take you no more than 20 minutes at this point) and I've been working on one weekend's worth of homework for a total of about 18 hours over the past two days, so I don't want to hear too many complaints about that. ;p
Okay, so moving on:
What is Number Theory exactly? A lot of you hit the nail on the... side of the head. What I mean is that you got the gist of it while missing some major points. Number theory IS the study of numbers and their behaviors, but it deals with a lot more than just sieves, lcms, gcds, prime numbers, and "harder" arithmetic. Number theory essentially connects everything together. You'll certainly be getting this feeling later on when dealing with things like the Chinese Remainder Theorem which is essentially a shortcut to solve systems of congruences when dealing with modular arithmetic (which should already clearly illustrate that this is a lot more than simple computation.) Furthermore, and especially recently, number theory has been applied in an ever-growing number of fields. Cryptosystems have long been a clever application of number theory and its principles, but more recently ideas have been applied to nearly all fields of applied mathematics (including physics, biology, chemistry, and even things like quantum computing and whatnot.) In a nutshell, there is a LOT you can do with number theory and the basic understanding of numbers doesn't hurt too much either.
For prompt number 2,
everyone was able to calculate the first 10 terms of the sequence (hopefully that is, I know some people who are far too quick to reach for their trusty ctrl+c/ctrl+v combo.) Unfortunately, it seems that no one was able to get the exact pattern without having seen it before. Certainly each number is the previous sum plus the current index (n,) but that's a tad too obvious. The closed form representation of the series on the other hand is a bit more elusive. Malerie, you came close by noting that n^2-n was 6 in the case where n=3. If you had gone a bit further with this you would've noticed that n^2+n was 12 in this case which happens to be twice the current sum. Furthermore, we can note that n^2+n is ALWAYS 2a_n (where a_n is the current sum.) A couple of seconds should allow everyone to convince themselves that this means that
(n^2+n)/2=a_n
That being said, if I want to know the sum of the first 1000000 positive integers, we simply plug in 1000000 and get
(1000000^2+1000000)/2
=(1000000000000+1000000)/2
=(1000001000000)/2
=500000500000
Now, how do we prove such a thing? There are a ton of ways, but two main ones that I'll mention. One is intuitive, the other is more mathematical. In the future if I ask you guys to prove (or try to prove) anything, you can go down either of these routes. Intuition can take you a long way.
Think of adding numbers like adding dots. We start with one dot, then add two, then three, and so on.
Let's consider the case that we have n=10, we can construct this picture:
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..
...
....
.....
......
.......
........
.........
..........
Clearly this pattern will continue on if we continue adding more dots. What we get is nothing but a triangle. Now let's try and find something we can calculate the area of easily. What's the easiest in this case? Squares/rectangles come to mind, so let's turn this into a square with red dots representing things we've just added.
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
As you can see, all this is is a mirror image of the number of dots we want to find. But what we have formed is a rectangle with 10 rows and 11 columns. What is the area of such a rectangle? Hopefully you're thinking to yourself "length*width" and if so, you're correct. So what is the area of the black part of the rectangle? Exactly half of that, right? So all we need is
length*width/2
In this case, the length is 10 and the width is 11 so there are 10*11/2=55 total black dots.
It shouldn't take long to convince yourself that this will hold for any number of rows. The length will always be n and the width will be n+1.
This means that our formula is a_n=n(n+1)/2
As for a more mathematical proof, let's write out what we know.
a_n=1+2+3+4+...+(n-2)+(n-1)+n
and if we change the order around
a_n=n+(n-1)+(n-2)+...+4+3+2+1
So we can add these two equations and get
2a_n=(n+1)+(n+1)+(n+1)+...+(n+1)
But clearly we only have n terms (so we have n of these (n+1)'s)
so that means we have 2a_n=n*(n+1)
or in other words:
a_n=n*(n+1)/2
So there are two different proofs. I should note that a lot of you mentioned you would just plug in a bunch of numbers and test it. This is NOT a reliable method of proof. I could find you infinitely many more formulas that "almost" work for this specific pattern but at some single point fail. Also, a lot of you mentioned that there must be a formula because there is a pattern. Ah, that would be nice wouldn't it? Unfortunately this is NOT always the case.
Consider the following sequence:
3,1,4,1,5,9,2,6,5,3,5,8,9,7,9,3,2,3,8,4,6,2,6,...
There is most definitely a pattern here, each term is the next digit in pi, or in other words that it is the next decimal place in the circumference of a circle divided by its diameter. Unfortunately, there is no algebraic formula for this sequence that does NOT involve the usage of pi itself. If anyone can come up with one I'd be happy to hear it (and claim whatever prize comes for breaking all of mathematics.) That said, do not assume that just because there is a pattern there must be a formula to it. A few people mentioned that "no one would want to sit and add all of those numbers." I can assure you that no one wants to calculate the one trillionth digit of pi by brute force and would LOVE to just plug in their number into a formula, but sometimes things just have to be done the hard way.
That's all for now, I'll be thinking of another question for next week. See you all then!
Ciao!
First things first, most everyone did fine with their responses (as like I mentioned, I was only asking for your points of view for these two. Do not expect these to stay QUITE as open ended as the year progresses.) I would like to note that while short answers ARE acceptable, I prefer to see the brief explanation to full paragraph length answers. This really isn't a lot of work to do once a week (it should take you no more than 20 minutes at this point) and I've been working on one weekend's worth of homework for a total of about 18 hours over the past two days, so I don't want to hear too many complaints about that. ;p
Okay, so moving on:
What is Number Theory exactly? A lot of you hit the nail on the... side of the head. What I mean is that you got the gist of it while missing some major points. Number theory IS the study of numbers and their behaviors, but it deals with a lot more than just sieves, lcms, gcds, prime numbers, and "harder" arithmetic. Number theory essentially connects everything together. You'll certainly be getting this feeling later on when dealing with things like the Chinese Remainder Theorem which is essentially a shortcut to solve systems of congruences when dealing with modular arithmetic (which should already clearly illustrate that this is a lot more than simple computation.) Furthermore, and especially recently, number theory has been applied in an ever-growing number of fields. Cryptosystems have long been a clever application of number theory and its principles, but more recently ideas have been applied to nearly all fields of applied mathematics (including physics, biology, chemistry, and even things like quantum computing and whatnot.) In a nutshell, there is a LOT you can do with number theory and the basic understanding of numbers doesn't hurt too much either.
For prompt number 2,
everyone was able to calculate the first 10 terms of the sequence (hopefully that is, I know some people who are far too quick to reach for their trusty ctrl+c/ctrl+v combo.) Unfortunately, it seems that no one was able to get the exact pattern without having seen it before. Certainly each number is the previous sum plus the current index (n,) but that's a tad too obvious. The closed form representation of the series on the other hand is a bit more elusive. Malerie, you came close by noting that n^2-n was 6 in the case where n=3. If you had gone a bit further with this you would've noticed that n^2+n was 12 in this case which happens to be twice the current sum. Furthermore, we can note that n^2+n is ALWAYS 2a_n (where a_n is the current sum.) A couple of seconds should allow everyone to convince themselves that this means that
(n^2+n)/2=a_n
That being said, if I want to know the sum of the first 1000000 positive integers, we simply plug in 1000000 and get
(1000000^2+1000000)/2
=(1000000000000+1000000)/2
=(1000001000000)/2
=500000500000
Now, how do we prove such a thing? There are a ton of ways, but two main ones that I'll mention. One is intuitive, the other is more mathematical. In the future if I ask you guys to prove (or try to prove) anything, you can go down either of these routes. Intuition can take you a long way.
Think of adding numbers like adding dots. We start with one dot, then add two, then three, and so on.
Let's consider the case that we have n=10, we can construct this picture:
.
..
...
....
.....
......
.......
........
.........
..........
Clearly this pattern will continue on if we continue adding more dots. What we get is nothing but a triangle. Now let's try and find something we can calculate the area of easily. What's the easiest in this case? Squares/rectangles come to mind, so let's turn this into a square with red dots representing things we've just added.
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
As you can see, all this is is a mirror image of the number of dots we want to find. But what we have formed is a rectangle with 10 rows and 11 columns. What is the area of such a rectangle? Hopefully you're thinking to yourself "length*width" and if so, you're correct. So what is the area of the black part of the rectangle? Exactly half of that, right? So all we need is
length*width/2
In this case, the length is 10 and the width is 11 so there are 10*11/2=55 total black dots.
It shouldn't take long to convince yourself that this will hold for any number of rows. The length will always be n and the width will be n+1.
This means that our formula is a_n=n(n+1)/2
As for a more mathematical proof, let's write out what we know.
a_n=1+2+3+4+...+(n-2)+(n-1)+n
and if we change the order around
a_n=n+(n-1)+(n-2)+...+4+3+2+1
So we can add these two equations and get
2a_n=(n+1)+(n+1)+(n+1)+...+(n+1)
But clearly we only have n terms (so we have n of these (n+1)'s)
so that means we have 2a_n=n*(n+1)
or in other words:
a_n=n*(n+1)/2
So there are two different proofs. I should note that a lot of you mentioned you would just plug in a bunch of numbers and test it. This is NOT a reliable method of proof. I could find you infinitely many more formulas that "almost" work for this specific pattern but at some single point fail. Also, a lot of you mentioned that there must be a formula because there is a pattern. Ah, that would be nice wouldn't it? Unfortunately this is NOT always the case.
Consider the following sequence:
3,1,4,1,5,9,2,6,5,3,5,8,9,7,9,3,2,3,8,4,6,2,6,...
There is most definitely a pattern here, each term is the next digit in pi, or in other words that it is the next decimal place in the circumference of a circle divided by its diameter. Unfortunately, there is no algebraic formula for this sequence that does NOT involve the usage of pi itself. If anyone can come up with one I'd be happy to hear it (and claim whatever prize comes for breaking all of mathematics.) That said, do not assume that just because there is a pattern there must be a formula to it. A few people mentioned that "no one would want to sit and add all of those numbers." I can assure you that no one wants to calculate the one trillionth digit of pi by brute force and would LOVE to just plug in their number into a formula, but sometimes things just have to be done the hard way.
That's all for now, I'll be thinking of another question for next week. See you all then!
Ciao!
Mal's Post 2
Part a.) What is the sum of the first n integers for n=1, 2, 3, 4, 5, 6, 7, 8, 9, and 10?
*Note, that is to say, for n=6 you will simply calculate 1+2+3+4+5+6
Answer:
1. 1
2. 3
3. 6
4. 10
5. 15
6. 21
7. 28
8. 36
9. 45
10. 55
Part b.) Is there a pattern as n gets larger? What do you intuitively think the pattern might be? If you don't see one immediately, try a few larger values of n (around 15 to 25.)
Answer:
You’re just adding n to the previous total. So each time it’s increasing by 1, 2, 3, 4, 5,…
Part c.) Do you think it's possible to write a formula for the sum of the first n integers for any n (for instance, if I want to know the sum of the first 1,000,000 natural numbers, but don't want to write all of them out, is there a shorter formula that I can simply plug n=1,000,000 into?) If so, what do you think it might be (or if you're not sure, why do you think there is one?) and if not, why don't you think so?
Answer:
Well, I’m not really positive. I feel like there is a formula…because I vaguely remember Tir telling me something….but I don’t know. But, I did find out that n^2-n gives you 6 for 3?
Part d.) If you think there is a formula, can you justify (or prove) that it is correct? If you don't think there is one, can you prove that there isn't? (It's okay if you can't do this part, but just elaborate a bit.)
Once again elaborating. Because there is a pattern, I believe there should be a formula. Isn’t that a given?
*Note, that is to say, for n=6 you will simply calculate 1+2+3+4+5+6
Answer:
1. 1
2. 3
3. 6
4. 10
5. 15
6. 21
7. 28
8. 36
9. 45
10. 55
Part b.) Is there a pattern as n gets larger? What do you intuitively think the pattern might be? If you don't see one immediately, try a few larger values of n (around 15 to 25.)
Answer:
You’re just adding n to the previous total. So each time it’s increasing by 1, 2, 3, 4, 5,…
Part c.) Do you think it's possible to write a formula for the sum of the first n integers for any n (for instance, if I want to know the sum of the first 1,000,000 natural numbers, but don't want to write all of them out, is there a shorter formula that I can simply plug n=1,000,000 into?) If so, what do you think it might be (or if you're not sure, why do you think there is one?) and if not, why don't you think so?
Answer:
Well, I’m not really positive. I feel like there is a formula…because I vaguely remember Tir telling me something….but I don’t know. But, I did find out that n^2-n gives you 6 for 3?
Part d.) If you think there is a formula, can you justify (or prove) that it is correct? If you don't think there is one, can you prove that there isn't? (It's okay if you can't do this part, but just elaborate a bit.)
Once again elaborating. Because there is a pattern, I believe there should be a formula. Isn’t that a given?
Trey Prompt #2
A.
1 = 1
2 = 3
3 = 6
4 = 10
5 = 15
6 = 21
7 = 28
8 = 36
9 = 45
10 = 55
B.
I am thinking that there is a pattern, you add the sum of what you got last time to the next number in number line.. I guess that is what you call it.
Example: 1+2=3 and then 3+3=6 and so on and so forth.
C.
I know that there is a formula but I cant seem to figure it out.
D.
I am positive that there is a formula but I am not sure that I can prove it correct.
1 = 1
2 = 3
3 = 6
4 = 10
5 = 15
6 = 21
7 = 28
8 = 36
9 = 45
10 = 55
B.
I am thinking that there is a pattern, you add the sum of what you got last time to the next number in number line.. I guess that is what you call it.
Example: 1+2=3 and then 3+3=6 and so on and so forth.
C.
I know that there is a formula but I cant seem to figure it out.
D.
I am positive that there is a formula but I am not sure that I can prove it correct.
Ryan's Post #2
a)
1 = 1
2 = 3
3 = 6
4 = 10
5 = 15
6 = 21
7 = 28
8 = 36
9 = 45
10 = 55
b) Honestly, the only pattern I see is just the next number is adding by 1, 2, 3, etc.
c) I think that there is a formula to this question, but I can't seem to figure it out.
I do think the formula would be something like the Fibonacci sequence formula, that if you have two consecutive numbers (ex. 15 and 21) you would be able to find the next number by adding 7, because you can figure that 21 - 15 = 6 so the next number would be 21 + 7.
d) I think that if there is a formula, it would have to have been proved if people use it.
Ryan
1 = 1
2 = 3
3 = 6
4 = 10
5 = 15
6 = 21
7 = 28
8 = 36
9 = 45
10 = 55
b) Honestly, the only pattern I see is just the next number is adding by 1, 2, 3, etc.
c) I think that there is a formula to this question, but I can't seem to figure it out.
I do think the formula would be something like the Fibonacci sequence formula, that if you have two consecutive numbers (ex. 15 and 21) you would be able to find the next number by adding 7, because you can figure that 21 - 15 = 6 so the next number would be 21 + 7.
d) I think that if there is a formula, it would have to have been proved if people use it.
Ryan
Dustin's Blog Prompt #1
Since i did prompt 2 last week, i'll do the other one this week. The question was, What is number theory? What i think that number theory is, is the theories and properties that allow formulas or patterns to occur. Number theory is analyzing the your patterns or sequences or series and finding something unique about it. Putting it into a formula if possible, or finding out why it occurs this way. Examples would be figuring out why an even number plus an even number is always and even number. Thats what I think number theory is.
Helen's Blog #2
A)
1:1
2:3
3:6
4:10
5:15
6:21
7:28
8:36
9:45
10:55
B) um.... I don't really know if there is a pattern because I don't really think I see one.... ummm...so yea :) haha
C) I think there is a formula because I doubt that nobody wants to sit down and calculate every single problem just to find this and it would take forever to finish.
D)I am sure there is a formula but I don't know if i'm able to prove it.
1:1
2:3
3:6
4:10
5:15
6:21
7:28
8:36
9:45
10:55
B) um.... I don't really know if there is a pattern because I don't really think I see one.... ummm...so yea :) haha
C) I think there is a formula because I doubt that nobody wants to sit down and calculate every single problem just to find this and it would take forever to finish.
D)I am sure there is a formula but I don't know if i'm able to prove it.
Feroy's Blog - Prompt 2
Yay. Blog. >_>
A.
1 = 1
2 = 3
3 = 6
4 = 10
5 = 15
6 = 21
7 = 28
8 = 36
9 = 45
10 = 55
B. There is pattern. What that pattern is, however, is beyond me. Well actually, I guess you could say that you can add the sum of the last value to the value being added to get the next value. If that makes sense.
Ex. First solution is 1, so 2 + 1 = 3, which is your next answer. 6 +4 = 10, so forth and so forth.
C. I'm pretty sure what I said can be translated into a formula, but I wouldn't be able to figure it out without looking it up. I mean, I get the concept, just don't know the formula.
D. Assuming that this is correct, the next two answers would be:
11 = 66
12 = 78
That's the best I can do. Not that good at proving things.
Justin #2
A)
1:1
2:3
3:6
4:10
5:15
6:21
7:28
8:36
9:45
10:55
B.) I do see a pattern here, and i think it is the sum of n + n-1, or in simpler terms, add the answer from the last n value. n=1 1+0=1 n=2 2+1=3 n=3 3+3... n=6 is 15+6=21, which is same as 1+2+3+4+5+6
C.) I think there is a formula for this, but I can't really explain it when it comes to relatively large numbers in the series.
D.) I can't exactly PROVE it, but I can kinda show that it is a type of sequence that we might be learning this week ha.
1:1
2:3
3:6
4:10
5:15
6:21
7:28
8:36
9:45
10:55
B.) I do see a pattern here, and i think it is the sum of n + n-1, or in simpler terms, add the answer from the last n value. n=1 1+0=1 n=2 2+1=3 n=3 3+3... n=6 is 15+6=21, which is same as 1+2+3+4+5+6
C.) I think there is a formula for this, but I can't really explain it when it comes to relatively large numbers in the series.
D.) I can't exactly PROVE it, but I can kinda show that it is a type of sequence that we might be learning this week ha.
Blog 2
ugh....really not feeling all that well, so this probably won't be "top notch" as you may say.....but here it goes
part a) the sum of the first n integers of a bunch of numbers and all that in which I don't know how to word this the right way and all that jazz? it sounds pretty easy to do, so I guess I'll get crackin.
n = 1 : 1
n = 2 : 3
n = 3 : 6
n = 4 : 10
n = 5 : 15
n = 6 : 21
n = 7 : 28
n = 8 : 36
n = 9 : 45
n = 10 : 55
part b) there's definitely a pattern. all you have to do is add whichever number of digits you have to the last number's sum. i.e. n=1 is 1, then n=2 is 3, which is basically like saying n=1+2, because you're just adding the main number you're using, it's pretty simple
part c) honestly, I have no idea at all. There probably is, and I wouldn't doubt that there's a formula way out there in the internet or something, but I have no idea what the formula probably is.
part d) whatever that formula may be, it's got to consist of something like n+x, where x is the sum of the past numbers' digits
part a) the sum of the first n integers of a bunch of numbers and all that in which I don't know how to word this the right way and all that jazz? it sounds pretty easy to do, so I guess I'll get crackin.
n = 1 : 1
n = 2 : 3
n = 3 : 6
n = 4 : 10
n = 5 : 15
n = 6 : 21
n = 7 : 28
n = 8 : 36
n = 9 : 45
n = 10 : 55
part b) there's definitely a pattern. all you have to do is add whichever number of digits you have to the last number's sum. i.e. n=1 is 1, then n=2 is 3, which is basically like saying n=1+2, because you're just adding the main number you're using, it's pretty simple
part c) honestly, I have no idea at all. There probably is, and I wouldn't doubt that there's a formula way out there in the internet or something, but I have no idea what the formula probably is.
part d) whatever that formula may be, it's got to consist of something like n+x, where x is the sum of the past numbers' digits
Kaitlyn's Blog #2
A) 1- 1
2- 3
3- 6
4- 10
5- 15
6- 21
7- 28
8- 36
9- 45
10- 55
B) Yes, i do see a pattern. To get the sum of the next integar n, you would simply take your sum, and add the next number to it. Ex: 6:21---so to find 7, you would add 7 to 6 to get 7:28.
C) I'm pretty sure there is a formula for this. Nobody wants to sit down and calculate every single problem just to find this. It would take to long. There is a pattern in the numbers, but I'm not sure what the formula would be for this.
D) I do think that there is a formula. If i knew the formula, I would test a few numbers to prove if it was the correct formula or not.
2- 3
3- 6
4- 10
5- 15
6- 21
7- 28
8- 36
9- 45
10- 55
B) Yes, i do see a pattern. To get the sum of the next integar n, you would simply take your sum, and add the next number to it. Ex: 6:21---so to find 7, you would add 7 to 6 to get 7:28.
C) I'm pretty sure there is a formula for this. Nobody wants to sit down and calculate every single problem just to find this. It would take to long. There is a pattern in the numbers, but I'm not sure what the formula would be for this.
D) I do think that there is a formula. If i knew the formula, I would test a few numbers to prove if it was the correct formula or not.
David's first Number Theory blog
Yay! Number Theory! Soooooo here we go I guess...
A.) n=10
1: 1
2: 3
3: 6
4: 10
5: 15
6: 21
7: 28
8: 36
9: 45
10: 55
B.) I don't know if there is a pattern but some of them are divisable by the numbers they were added from.
C.) There probably is a formula for this because there are formulas for everything and you probably would not have included this in the prompt if one didn't exist. I haven't the slightest idea as to what it would be though.
D.) I think a formula would work because there are rules to addition and all of those other functions and whatnot so that a pattern probably does eventually emerge and one could use it to figure it out without adding 1,000,000+ numbers using case work (any written or shown work) or otherwise.
A.) n=10
1: 1
2: 3
3: 6
4: 10
5: 15
6: 21
7: 28
8: 36
9: 45
10: 55
B.) I don't know if there is a pattern but some of them are divisable by the numbers they were added from.
C.) There probably is a formula for this because there are formulas for everything and you probably would not have included this in the prompt if one didn't exist. I haven't the slightest idea as to what it would be though.
D.) I think a formula would work because there are rules to addition and all of those other functions and whatnot so that a pattern probably does eventually emerge and one could use it to figure it out without adding 1,000,000+ numbers using case work (any written or shown work) or otherwise.
post number two
a.1-1
2-3
3-6
4-10
5-15
6-21
7-28
8-36
9- 45
10-55
b. um, i'm just gonna be straight up with you and say i don't see a pattern at all... haha. and i thought about looking it up to see if there was one, but i decided i'd just tell you the truth.
c. yes, there probably is a formula. because i personally think that there is a formula for almost everything in the mathematical world.. i may be wrong. but that's just my opinion. no, i don't know the formula, because i'm not very good at looking at things and being able to describe why it is that why.. i have a very strange train of thought that not many people would understand.
d. yes, i think there's a formula. because each number is adding the same amout to it, plus one more than the previous time.. now do i know how to make a formula out of that? nope. ha.
2-3
3-6
4-10
5-15
6-21
7-28
8-36
9- 45
10-55
b. um, i'm just gonna be straight up with you and say i don't see a pattern at all... haha. and i thought about looking it up to see if there was one, but i decided i'd just tell you the truth.
c. yes, there probably is a formula. because i personally think that there is a formula for almost everything in the mathematical world.. i may be wrong. but that's just my opinion. no, i don't know the formula, because i'm not very good at looking at things and being able to describe why it is that why.. i have a very strange train of thought that not many people would understand.
d. yes, i think there's a formula. because each number is adding the same amout to it, plus one more than the previous time.. now do i know how to make a formula out of that? nope. ha.
Stephen's blog--Prompt 2
a)
1 - 1
2 - 3
3 - 6
4 - 10
5 - 15
6 - 21
7 - 28
8 - 36
9 - 45
b) As n gets larger, the value that comes out is the previous value added to n. The pattern is very uniform and easy to figure out, to me at least.
c) I think there is a formula, but I do not think I am simply smart enough to think of it haha. All I know is to get another number, you add the previous number to the n integer you are trying to find at the moment, if that makes sense? I don't know.
d) Again, I do believe there is a formula and exists, because I think we learned this in Probability/Statistics last year and/or Advanced Math. I think Jordan is good with this kind of stuff...I supposed you would ask something like this Jordan, thanks. But I really don't remember anything at the moment.
Alaina's blog, 12 Sept 2010
A)
Yes, there is most definitely a pattern. I provided the work that shows the pattern the sum of integers [1,2,3,4,5,6,7,8,9,10] follows. The first integer will follow numerical order and the second will be the sum of previous integers.
C/D)
It is possible to have a formula for this type of problem. First, a reason for having a formula for this would be because NO ONE wants to sit down and add all of the integers up to 1,000,000. Since there is a common pattern to arrive at the next integer, the has to be a formula. However, some series do not follow any pattern and, therefore, do not have a formula. The formula that this particular series follows would be [n(n+1)/2] only having to plug in for "n". If you plugged in 1,000,000- [1,000,000(1,000,000+1)/2]=[1,000,000(1,000,001)]/2= 500,000,500,000.This formula proves accurate for natural and positive integers.
- n=1-1:1 to get next integer n=2 1+(n=2)
- n=2-2:3 to get next integer n=3 2+1+(n=3)
- n=3-3:6 to get next integer n=4 3+2+1+(n=4)
- n=4-4:10 to get next integer n=5 4+3+2+1+(n=5)
- n=5-5:15 to get next integer n=6 5+4+3+2+1+(n=6)
- n=6-6:21 to get next integer n=7 6+5+4+3+2+1+(n=7)
- n=7-7:28 to get next integer n=8 7+6+5+4+3+2+1+(n=8)
- n=8-8:36 to get next integer n=9 8+7+6+5+4+3+2+1+(n=9)
- n=9-9:45 to get next integer n=10 9+8+7+6+5+4+3+2+1+(n=10)
- n=10-10:55
Yes, there is most definitely a pattern. I provided the work that shows the pattern the sum of integers [1,2,3,4,5,6,7,8,9,10] follows. The first integer will follow numerical order and the second will be the sum of previous integers.
C/D)
It is possible to have a formula for this type of problem. First, a reason for having a formula for this would be because NO ONE wants to sit down and add all of the integers up to 1,000,000. Since there is a common pattern to arrive at the next integer, the has to be a formula. However, some series do not follow any pattern and, therefore, do not have a formula. The formula that this particular series follows would be [n(n+1)/2] only having to plug in for "n". If you plugged in 1,000,000- [1,000,000(1,000,000+1)/2]=[1,000,000(1,000,001)]/2= 500,000,500,000.This formula proves accurate for natural and positive integers.
Mary prompt 1 and 2
I just got in this blog, it finally worked, so I hope I'll get credit for both the prompts...
Prompt 1
I think number theory is maybe, how numbers came to be perhaps? Or a theory on how to work any and every type of problem? I really have no idea. Some secret sequences for working big problems that no one else knows about? So, I guess I'll find out soon.
Prompt 2
1 : 1
2 : 3
3 : 6
4 : 10
5 : 15
6 : 21
7 : 28
8 : 36
9 : 45
10 : 55
B) pattern? what do i look like? i have no clue. haha, i don't know if this counts for anything, but if you look at the diagonal numbers and add them...they give you the number on the right. but that still means you would have to write this out for a huge number which probably wouldn't help...
C) David sitting right over here, told me that there is a formula, but I don't know what it is.
D) So I know there is a formula; how would i prove that is it correct? you could type it in your calculator, then use the formula to check yourself i guess..
byebye.
Prompt 1
I think number theory is maybe, how numbers came to be perhaps? Or a theory on how to work any and every type of problem? I really have no idea. Some secret sequences for working big problems that no one else knows about? So, I guess I'll find out soon.
Prompt 2
1 : 1
2 : 3
3 : 6
4 : 10
5 : 15
6 : 21
7 : 28
8 : 36
9 : 45
10 : 55
B) pattern? what do i look like? i have no clue. haha, i don't know if this counts for anything, but if you look at the diagonal numbers and add them...they give you the number on the right. but that still means you would have to write this out for a huge number which probably wouldn't help...
C) David sitting right over here, told me that there is a formula, but I don't know what it is.
D) So I know there is a formula; how would i prove that is it correct? you could type it in your calculator, then use the formula to check yourself i guess..
byebye.
Post # 2
Whats up guysssssssssss.
A.) 1: 1
2: 3
3: 6
4: 10
5: 15
6: 21
7: 28
8: 36
9: 45
B.) i'm not sure what the pattern is, but they are all divisable by the number added to them..
C.) If i had to pick a formula it would be something like n(n + n+1) or something like that. It has to be something saying taht there is a divisor in there somewhere..
D.) I do think there is a formula that can interpret the pattern of this. It would work because every number has certain qualities involving certain addition, subtraction, multiplication, and division rules.
A.) 1: 1
2: 3
3: 6
4: 10
5: 15
6: 21
7: 28
8: 36
9: 45
B.) i'm not sure what the pattern is, but they are all divisable by the number added to them..
C.) If i had to pick a formula it would be something like n(n + n+1) or something like that. It has to be something saying taht there is a divisor in there somewhere..
D.) I do think there is a formula that can interpret the pattern of this. It would work because every number has certain qualities involving certain addition, subtraction, multiplication, and division rules.
Saturday, September 11, 2010
Week 2 Blog Prompt
So for those of you who haven't seen (or have forgotten,) here is a repost of the prompt for this week:
Part a.) What is the sum of the first n integers for n=1, 2, 3, 4, 5, 6, 7, 8, 9, and 10?
*Note, that is to say, for n=6 you will simply calculate 1+2+3+4+5+6
Part b.) Is there a pattern as n gets larger? What do you intuitively think the pattern might be? If you don't see one immediately, try a few larger values of n (around 15 to 25.)
Part c.) Do you think it's possible to write a formula for the sum of the first n integers for any n (for instance, if I want to know the sum of the first 1,000,000 natural numbers, but don't want to write all of them out, is there a shorter formula that I can simply plug n=1,000,000 into?) If so, what do you think it might be (or if you're not sure, why do you think there is one?) and if not, why don't you think so?
Part d.) If you think there is a formula, can you justify (or prove) that it is correct? If you don't think there is one, can you prove that there isn't? (It's okay if you can't do this part, but just elaborate a bit.)
Keep in mind that you are not expected to do any research (and in fact, it looks worse if you do.) It will be relatively obvious if you use an outside source or another student's blog post as a basis for your own.
Part a.) What is the sum of the first n integers for n=1, 2, 3, 4, 5, 6, 7, 8, 9, and 10?
*Note, that is to say, for n=6 you will simply calculate 1+2+3+4+5+6
Part b.) Is there a pattern as n gets larger? What do you intuitively think the pattern might be? If you don't see one immediately, try a few larger values of n (around 15 to 25.)
Part c.) Do you think it's possible to write a formula for the sum of the first n integers for any n (for instance, if I want to know the sum of the first 1,000,000 natural numbers, but don't want to write all of them out, is there a shorter formula that I can simply plug n=1,000,000 into?) If so, what do you think it might be (or if you're not sure, why do you think there is one?) and if not, why don't you think so?
Part d.) If you think there is a formula, can you justify (or prove) that it is correct? If you don't think there is one, can you prove that there isn't? (It's okay if you can't do this part, but just elaborate a bit.)
Keep in mind that you are not expected to do any research (and in fact, it looks worse if you do.) It will be relatively obvious if you use an outside source or another student's blog post as a basis for your own.
Heather's #2
:D YESSSSSSSSSSS! I know how to do this because Alex taught me this for my sequences and series test at nationals!! Ha anywayyyy this is what you would get using this method > (*n=6 >> 1+2+3+4+5+6)
A.)
n=1: 1
n=2: 3
n=3: 6
n=4: 10
n=5: 15
n=6: 21
n=7: 28
n=8: 36
n=9: 45
n=10: 55
B.) Yes there is definitely a pattern. First if you look at the row of sums, you notice that you get the next number by adding the next "n"...As in if you take 1, your first sum, and add 2 you get your second sum which is 3. Adding 3 to your second sum you get 6, your third sum. Then adding 4 to 6 you get your 4th sum which is 10 and so on...So in other words (looking at part A) you would add the two red numbers to get the blue number. Also, looking more closely at the sums, I notice that each n is either a divisor of or has a divisor of the sum number. For instance, for n=7, 7 is a divisor of 28..meaning that 7 can go into 28 evenly. Also, for n=4, 2 is a divisor of both 4 and 10 since 4 cannot go into 10 evenly.
C./D.) Yes it is possible to have a formula for this type of problem...but seeing as how I learned it already, I'll explain why I think there is a formula for this. First of all, I see a reason for having a formula for this because we all know that no one wants to sit down and add up all the numbers up to 1,000,000 and surely no mathematician had the time for that either. I assumed there had to be a formula since there is a common pattern here. No matter how large the sum becomes, the pattern is still there. However, that isn't always the case, but I figured there must be a pattern here since the sums are in ascending order. Sooo with that, the formula I learned is n(n+1)/2. (I honestly did learn this, I didn't look it up) To use it, all you have to do is plug in n. So if I really do want to find the sum of the first 1,000,000 natural numbers then I would get 1,000,000(1,000,001)/2 which would simplify to 500000500000. I'd say this formula is pretty accurate for natural numbers/positive integers specifically; however I'm not sure if it would work the same say if you were asked to find the sum of the first 100 negative integers..
A.)
n=1: 1
n=2: 3
n=3: 6
n=4: 10
n=5: 15
n=6: 21
n=7: 28
n=8: 36
n=9: 45
n=10: 55
B.) Yes there is definitely a pattern. First if you look at the row of sums, you notice that you get the next number by adding the next "n"...As in if you take 1, your first sum, and add 2 you get your second sum which is 3. Adding 3 to your second sum you get 6, your third sum. Then adding 4 to 6 you get your 4th sum which is 10 and so on...So in other words (looking at part A) you would add the two red numbers to get the blue number. Also, looking more closely at the sums, I notice that each n is either a divisor of or has a divisor of the sum number. For instance, for n=7, 7 is a divisor of 28..meaning that 7 can go into 28 evenly. Also, for n=4, 2 is a divisor of both 4 and 10 since 4 cannot go into 10 evenly.
C./D.) Yes it is possible to have a formula for this type of problem...but seeing as how I learned it already, I'll explain why I think there is a formula for this. First of all, I see a reason for having a formula for this because we all know that no one wants to sit down and add up all the numbers up to 1,000,000 and surely no mathematician had the time for that either. I assumed there had to be a formula since there is a common pattern here. No matter how large the sum becomes, the pattern is still there. However, that isn't always the case, but I figured there must be a pattern here since the sums are in ascending order. Sooo with that, the formula I learned is n(n+1)/2. (I honestly did learn this, I didn't look it up) To use it, all you have to do is plug in n. So if I really do want to find the sum of the first 1,000,000 natural numbers then I would get 1,000,000(1,000,001)/2 which would simplify to 500000500000. I'd say this formula is pretty accurate for natural numbers/positive integers specifically; however I'm not sure if it would work the same say if you were asked to find the sum of the first 100 negative integers..
Tuesday, September 7, 2010
Ryan's #1
What is number theory?
When I hear number theory, I think of the branch of math that explains how numbers work. I think of the properties of how numbers and patterns work. I think it is about sequences, series, modular numbers, and other subjects dealing with how numbers work.
ryanbreauddawg(:)
When I hear number theory, I think of the branch of math that explains how numbers work. I think of the properties of how numbers and patterns work. I think it is about sequences, series, modular numbers, and other subjects dealing with how numbers work.
ryanbreauddawg(:)
Monday, September 6, 2010
Mal's Post 1
What is number theory?
Okay. Simple right? I believe number theory is a branch of mathematics that deals with just what the title implies...Numbers. Imagine that. In a way, it could be viewed as a higher level arithmetic considering much of the coursework involves studying numbers and manipulating them. Come to think of it, number theory is also the branch of mathematics that deals with the properties of integers, rational, natural etc. numbers...
According to the Mathematical Atlas (see link below), there are many sub-branches, if you will, of number theory. The way I look at it is the following: little kid number theory and big kid number theory. Remember when you were in 4th grade going over, for what probably was the thousandth time, prime and composite numbers? yeah, number theory. Now think back to the first/second week of school. What were we doing? Prime and composite numbers correct? Except, that sieve of eranth(whatever) was NOT something we used in 4th grade. Hence, the big kid title.
Kthnxbai...:DD
ohh...almost forgot. Source= this website: http://www.math.niu.edu/~rusin/known-math/index/11-XX.html
Okay. Simple right? I believe number theory is a branch of mathematics that deals with just what the title implies...Numbers. Imagine that. In a way, it could be viewed as a higher level arithmetic considering much of the coursework involves studying numbers and manipulating them. Come to think of it, number theory is also the branch of mathematics that deals with the properties of integers, rational, natural etc. numbers...
According to the Mathematical Atlas (see link below), there are many sub-branches, if you will, of number theory. The way I look at it is the following: little kid number theory and big kid number theory. Remember when you were in 4th grade going over, for what probably was the thousandth time, prime and composite numbers? yeah, number theory. Now think back to the first/second week of school. What were we doing? Prime and composite numbers correct? Except, that sieve of eranth(whatever) was NOT something we used in 4th grade. Hence, the big kid title.
Kthnxbai...:DD
ohh...almost forgot. Source= this website: http://www.math.niu.edu/~rusin/known-math/index/11-XX.html
Sunday, September 5, 2010
What is Number Theory???
Number Theory is a branch of mathematics that studies intergers and their relation and properties to each other. It deals with different classifications that intergers and numbers can be grouped into. It is basically an easier way to solving complicating math processes by using easier steps and easy to grasp methods
Blog 1
Wow, what is number theory?
Pshh, that's an easy question.
It's......damn, forgot........
I'm just gonna wing it right here, so don't mind me. :P
What do I think number theory is?
It's the theory of numbers, duh.
I do remember something about cracking codes though. That sounds really fun. *sarcasm*
I seriously have no idea....
*bursts into a rap*
Yo this is a story all about how numbers get theorized into oblivion.
They got codes being cracked, alarms being set off, and a whole buncha dogs chasing after us.
So then I heard ma dawgs talkin bout how this kind of theory can help us get away with this one.
They said "If we hit this jump at the right angle, we gonna get out of this joint."
I told em that I couldn't figure it out too easily.
They questioned me and then they told this to me:
"Boy you don't get it? Man go take this class. It's gonna help you out in all kinds of fields like that.
And if you actually do good, they gonna take you in the math club, which is mandatory in that class man."
I questioned em with such a surprised look and asked em "Dudes, how do yall even know about this class."
They didn't tell me and just sent me here.
Now here I am, in this class, not knowing a thing about Number Theory.
It'll be a while till I figure it out.
Pshh, that's an easy question.
It's......damn, forgot........
I'm just gonna wing it right here, so don't mind me. :P
What do I think number theory is?
It's the theory of numbers, duh.
I do remember something about cracking codes though. That sounds really fun. *sarcasm*
I seriously have no idea....
*bursts into a rap*
Yo this is a story all about how numbers get theorized into oblivion.
They got codes being cracked, alarms being set off, and a whole buncha dogs chasing after us.
So then I heard ma dawgs talkin bout how this kind of theory can help us get away with this one.
They said "If we hit this jump at the right angle, we gonna get out of this joint."
I told em that I couldn't figure it out too easily.
They questioned me and then they told this to me:
"Boy you don't get it? Man go take this class. It's gonna help you out in all kinds of fields like that.
And if you actually do good, they gonna take you in the math club, which is mandatory in that class man."
I questioned em with such a surprised look and asked em "Dudes, how do yall even know about this class."
They didn't tell me and just sent me here.
Now here I am, in this class, not knowing a thing about Number Theory.
It'll be a while till I figure it out.
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