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?
Sunday, September 26, 2010
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...
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