Remainder Prob

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by asax » Wed Jun 20, 2012 1:57 am
A :) First thought it might be E. But then only noticed its n^2 minus 1 !! he he :)
Looking forward to 2013 MBA admissions!

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by UmangMathur » Wed Jul 11, 2012 4:34 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.
The only exception to this is 1. But that won't trouble the answer in any way, because in that case, the product will be 0 - (1-1)*(1+1) = 0 thus choice A will hold true.

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by hah177 » Wed Jul 11, 2012 9:06 am
1. If n is odd integer, n = 2k + 1
We have : n^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k = 4k(k + 1) is multiple of 8. So r = 0 : Sufficient ---> AD

2. If n = 1 : r = 0
If n = 2 : r = 3
Hence, (2) is insufficient ---> A is correct anwser.

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by Shubhu@MBA » Thu Jul 19, 2012 11:13 am
For the expression (n^2)-1/8

If n is odd, remainder will always be 0

for e.g. 3^2 -1=9-1=8/8. Rem =0
5^2-1=24/8=3. Rem =0

13^2=169-1=168/8/. Rem=0

Thus option 1 is sufficient to answer the question.

Coming to Option 2,
2. n is not divisible by 8

This is not sufficient as n can be any number.

for e.g (9^2) -1=80/8. Rem =0
(10^2) -1=99/8. Rem =3
Thus option 2 cannot provide us the correct answer for the question.

Hence, option 1 is the answer.

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by rideema » Fri Jul 20, 2012 10:54 pm
Dear mana6,

while accessing the first statement you are taking even values for n, whereas the statement states that n is odd.

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by brijgmat » Fri Jan 04, 2013 10:25 am
Answer is A.

1. If n is odd number, it can be 3, 5, 7, 9, 11, ...
so n^2-1 will be 8, 24, 48, 80, 120,...all are multiple of 8. Sufficient

2. For n = 2, 3...r will be 3, 0 Insufficient

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by davidschneider » Sun Jan 06, 2013 6:00 am
Only statement number 1 is sufficient.

Answer: A

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by The Iceman » Tue Jan 08, 2013 5:03 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
If n is odd it can be expressed as 2k+1 for any integer k.
So, n^2-1 = (2k+1)^2-1 =4k^2+4k=4k(k+1)=8m (where m is another integer)=> r=0 for all odd n.

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by rajeshsinghgmat » Wed Jan 16, 2013 10:41 pm
A for answer.

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by ritzzzr » Mon Jan 21, 2013 1:50 am
Ans is A

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by vongochao » Tue Feb 05, 2013 12:38 am
just 10 second to answer : A.
n^2 always even integer
so n^2- 1 is an odd.
odd/8(even integer)= odd No.

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by aceacharya » Fri Mar 08, 2013 2:39 am
The majority is right
The answer is A

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by gaurav_gaur » Sun Apr 21, 2013 11:41 am
Clearly `a` is the answer

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by reubenv18 » Wed Jun 26, 2013 8:14 am
Why aren't the answer choices listed? How am I supposed to choose an answer if infant see the choices? Please help!!!!

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by nguyenduong » Sat Jul 20, 2013 1:18 am
a. if n is odd then n must be 1,3, 5,7,9 then the last digit of n^2 and (n^2-1) must be 1,9,5,9 and 0,8,4,8.
Let (n^2-1) / 8 means that (n^2-1): 2:2:2.
we have the number with last digit 8 is always divisible by 2 ( for last digit only 8:2=4:2=2:2=1) ; the number with last digit 0 will have remainder 1 after dividing by 2:2:2, the number with last digit 4 will have remainder 1 after dividing by 2:2:2.
then the remander will be always 1, A is sufficient
b.n is not divisible by 8 then n can be odd or even => B is insufficient.

the answer is A