## Remainder Prob

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### Remainder Prob

by zagcollins » Mon Jul 28, 2008 9:10 am
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
Last edited by zagcollins on Tue Jul 29, 2008 4:31 am, edited 1 time in total.

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by augusto » Mon Jul 28, 2008 12:11 pm
IMHO the answer is A.

1) Just plug in odd numbers,
n=1 : (1^2 - 1)/8 = 0
n=3 : (3^2 - 1)/8 = 0
n=5 : (5^2 - 1)/8 = 0
n=7 : (7^2 - 1)/8 = 0
after this I assume that r is always 0, so it´s sufficient

2) Again plug some numbers
n=1 : (1^2 - 1)/8 = 0
n=2 : (2^2 - 1)/8 = 3
upz, not sufficient.
Last edited by augusto on Mon Jul 28, 2008 12:21 pm, edited 1 time in total.

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by ricky » Mon Jul 28, 2008 12:16 pm
IMO A.

R is always 0

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### Re: Remainder Prob

by Ian Stewart » Tue Jul 29, 2008 5:47 am
zagcollins 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
You might notice that this is a difference of squares:

n^2 - 1 = (n+1)(n-1)

If n is odd, then n-1 and n+1 are consecutive even integers. If you take any two consecutive even integers, one of them will be divisible by 4, the other not, so their product must be divisible by 8. Thus, if we know n is odd, we can be certain that n^2 -1 will be divisible by 8, and r will be zero. 1) is sufficient. 2) is not; n might be even, or might be odd. A.
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by sanjaylakhani » Sat Sep 06, 2008 10:26 am
but what if N=1, which is possible since it is a positive odd integer, then remainder is 8 -right

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by Ian Stewart » Sat Sep 06, 2008 10:40 am
sanjaylakhani wrote:but what if N=1, which is possible since it is a positive odd integer, then remainder is 8 -right
When you divide any integer by 8, the only possible remainders are:

0, 1, 2, 3, 4, 5, 6 and 7

The remainder can never be 8 when you divide by 8. In addition, 0 is divisible by every positive integer, so the remainder is 0 when you divide 0 by 8 (or by any other positive integer).
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by sanjaylakhani » Sat Sep 06, 2008 11:03 am
Thanks a lot stewart...either I am sleepy or plain stupid ... ...I am embarrased....

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### Re: Remainder Prob

by anniev2 » Wed Feb 18, 2009 8:17 pm
Ian Stewart wrote:If you take any two consecutive even integers, one of them will be divisible by 4, the other not, so their product must be divisible by 8.
Thanks for the great explanation. Can you please explain this a little further. I am trying to understand the concept rather than just the answer to this question.

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by cramya » Wed Feb 18, 2009 8:27 pm
Anniev,

Take any 2 consecutive even numbers & u will notice:

1) One of the numbers will always be divisible by 4
2)The product will always be divisible by 8 since one number is alreayd divisible by 4 and the other being even will always provide 2 as a factor

Hope this helps!

Regards,
CR

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by anniev2 » Thu Feb 19, 2009 5:03 am
cramya wrote:Anniev,

Take any 2 consecutive even numbers & u will notice:

1) One of the numbers will always be divisible by 4
2)The product will always be divisible by 8 since one number is alreayd divisible by 4 and the other being even will always provide 2 as a factor

Hope this helps!

Regards,
CR
Yes, this is helpful...thank you.

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by arzanr » Sat Mar 27, 2010 10:18 am
Ian Stewart wrote:
zagcollins 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
You might notice that this is a difference of squares:

n^2 - 1 = (n+1)(n-1)

If n is odd, then n-1 and n+1 are consecutive even integers. If you take any two consecutive even integers, one of them will be divisible by 4, the other not, so their product must be divisible by 8. Thus, if we know n is odd, we can be certain that n^2 -1 will be divisible by 8, and r will be zero. 1) is sufficient. 2) is not; n might be even, or might be odd. A.
This is what I call mathemajics!

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by akahuja143 » Sat Mar 27, 2010 10:27 am
Thanks for sharing the cool tip Ian

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by Abhishek009 » Wed Nov 10, 2010 2:47 am
zagcollins 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
Only statement 1 is sufficient:

Checked with the following as n :

n = 3

n = 5

n = 7

However we can't arrive the answer using statement 2 alone.
Abhishek

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by vorand » Fri Nov 12, 2010 11:49 am
n - odd => n=2k+1, k=0,1,2,...

n^2-1= (n+1)(n-1) = (2k+2)*2k= 4k*(k+1)

4k*(k+1)/8=1/2*k*(k+1) => k or k+1 is even => remainder (n^2-1)/8 is 0

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by mana6 » Sat Jan 08, 2011 3:19 pm
If n is a positive integer and r is the remainder when (n^2)-1 divided by 8 ,what is the value of r?

I do not understand how the answer is a. i did this to get e as my answer:

1) n is not odd

n q r
4 1 7
6 4 3
8 7 7

This statement is insufficient because it doesnt definitively give you one answer for r. or so i thought

2) n is not divisible by 8

n q r
12 17 7
10 12 3
7 6 0

this again doesnt definitively give you one value for r so i said this was also insufficient.

1and 2) n is not odd and it is not divisible by 8

n q r n q r
2 0 3 6 4 3
4 1 7 10 12 3

when i tried twelve it also gave me a remainder of seven so since r can be three or seven this is also not enough. is there something im missing?