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If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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11 Feb 2010, 14:52
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If n is a positive integer and r is the remainder when n^2  1 is divided by 8, what is the value of r? (1) n is odd (2) n is not divisible by 8 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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Last edited by Bunuel on 28 Jul 2015, 06:52, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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11 Feb 2010, 15:06
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given, n is positive interger and n^2  1 = 8 * k + r > r remainder what is r??
st 1) n is odd n^21 = (n+1) * (n1) so n+1 and n1 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: If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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11 Feb 2010, 15:19
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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)(n1). 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: If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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11 Feb 2010, 16:11
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: If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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11 Feb 2010, 16:18
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: If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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11 Feb 2010, 16:26
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: If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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15 Feb 2010, 01:45
given n is postive integer n^21/8
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



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Re: If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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23 Aug 2016, 21:03
YTT wrote: If n is a positive integer and r is the remainder when n^2  1 is divided by 8, what is the value of r? (1) n is odd (2) n is not divisible by 8 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!! Please check the solution in the file attached
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Re: If n is a positive integer and r is the remainder when n^2  1 is divi [#permalink]
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08 Oct 2016, 08:55
YTT wrote: If n is a positive integer and r is the remainder when n^2  1 is divided by 8, what is the value of r? (1) n is odd (2) n is not divisible by 8 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!! n^21=(n+1)(n1) (1) when n is odd, given expression is always mutiple of 8 hence zero(0) remainder Sufficient (2) put the various nos. except multiple of 8 and you will get various remainder unsufficient




Re: If n is a positive integer and r is the remainder when n^2  1 is divi
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08 Oct 2016, 08:55







