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# 700 Algrbra! Need help again. Thanks so much!

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700 Algrbra! Need help again. Thanks so much! [#permalink]  11 Feb 2010, 13:52
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Could anyone explain to me why the number 1 would work in the first situation? I understand why 3, 5, 7 or others work. But why 1 works, too? Thank you so much for this great help!!
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Re: 700 Algrbra! Need help again. Thanks so much! [#permalink]  11 Feb 2010, 14:06
given, n is positive interger and
n^2 - 1 = 8 * k + r -> r remainder
what is r??

st 1) n is odd
n^2-1 = (n+1) * (n-1)
so n+1 and n-1 are consequetive even numbers... one of them will be multiple of 2 and the other will be multiple of 4. So n^2 - 1 will be evenly divided by 8 and r=0
Sufficient
st 1) n is not divisible by 8. Not sufficient

A
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Re: 700 Algrbra! Need help again. Thanks so much! [#permalink]  11 Feb 2010, 14:19
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I like the approach in the post above best for this problem; when you see something like n^2 - 1 in a GMAT question, it will almost always be useful to use the difference of squares factorization: n^2 - 1 = (n+1)(n-1). A less elegant alternative is to write n = 2k + 1. Then n^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 + 4k = 4(k)(k + 1), and since k and k+1 are consecutive integers, one of them must be divisible by 2, so 4(k)(k + 1) must be divisible by 4*2 = 8.

To answer the question in the original post, if n=1, then n^2 - 1 = 0. So the question becomes, what is the remainder when 0 is divided by 8? Well, 0 is divisible by every positive integer; the quotient is zero and the remainder is zero. If you think back to how you first learned division, this should hopefully be clear: if you have, say, 11 apples and 8 children, we can give each child 1 apple and we have 3 left over, so the quotient is 1 and the remainder is 3 when you divide eleven by eight. If we have 0 apples and 8 children, we can give each child 0 apples and we have 0 left over, so the quotient is 0 and the remainder is 0 when we divide zero by eight.
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Re: 700 Algrbra! Need help again. Thanks so much! [#permalink]  11 Feb 2010, 15:11
Thanks for the great help.

But I still feel very confused.
The question is asking "what's the value of r"?
I understand when n is 3,5,7, the r will be 1. However, if n is 1, r will be 0. In this case, we have two answers for r and we can't really tell the exact value for r, right? This is the reason why I don't think the first one work and the answer should be "E". Am I in the right path? Thanks for the help again.
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Re: 700 Algrbra! Need help again. Thanks so much! [#permalink]  11 Feb 2010, 15:18
Expert's post
YTT wrote:
Thanks for the great help.

But I still feel very confused.
The question is asking "what's the value of r"?
I understand when n is 3,5,7, the r will be 1. However, if n is 1, r will be 0. In this case, we have two answers for r and we can't really tell the exact value for r, right? This is the reason why I don't think the first one work and the answer should be "E". Am I in the right path? Thanks for the help again.

No, the remainder will be zero for any of the values 1, 3, 5, or 7 (or for any other odd value of n). If, say, n=5, then n^2 - 1 = 25 - 1 = 24, and the remainder when we divide 24 by 8 is zero.
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Re: 700 Algrbra! Need help again. Thanks so much! [#permalink]  11 Feb 2010, 15:26
Yes, You are right! Sorry that I forgot that we have to "-1". Thank you so much for this!! Now, I got it!
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Re: 700 Algrbra! Need help again. Thanks so much! [#permalink]  15 Feb 2010, 00:45
given n is postive integer n^2-1/8

1. n is odd

if n is odd then the values goes 1,3,5,7,...

if n=1 thn 0/8 = 0 so remainder is 0
if n=3 then 8/8= 1 so remanider is 0
if n=5 then 34/8 = 3 so remainder is 0

so clearly A is sufficient
Re: 700 Algrbra! Need help again. Thanks so much!   [#permalink] 15 Feb 2010, 00:45
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