Remainder Problem

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Remainder Problem

by Akansha » Thu May 26, 2011 9:40 am
If n is a positive integer and r is remainder when (n-1)(n+1) is divided by
24, what is the value of r?
a. n is divisible by 2
b. n is not divisible by 3


OA is C

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by clock60 » Thu May 26, 2011 10:37 am
hi Akansha
it must be a mistake in statements, or in oa
with given info the answer must be E
(1) says that n is even
n=2, 1*3=3=24*0+3 remainder 3
n=4, 3*5=15=24*0+15 remaider 15
not suff
(2)n is not divisible by 3
again example from 1 st, n=4, remaider 15
n=10, 9*11=99=24*4+3 remaider is 3
not suff
both n is even and not divisible by 3
n=4 remaider 15
n=10 remaider 3
it comes that the answer is E

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by Anurag@Gurome » Thu May 26, 2011 8:31 pm
Akansha wrote:If n is a positive integer and r is remainder when (n-1)(n+1) is divided by
24, what is the value of r?
a. n is divisible by 2
b. n is not divisible by 3


OA is C

Statement 1) should read as 'n is not divisible by 2'.
Solution:
Let us first consider statement 1) alone.
If we take n = 5 and n = 9, we will get two different values for r.
Or 1) alone is not sufficient to answer the question.
We next consider 2) alone.
If we take n = 2 and n = 4, we get two different values for r.
Or 2) alone is not sufficient to answer the question.
Next, combine both the statements together and check.
On combining, we have that n is of the form 6k+1 or 6k+5, k being 0,1,2,3....
If n = 6k+1, (n+1)(n-1) = (6k+2)(6k)= 12k(3k+1). If k is even, 12k is divisible by 24 and hence n = 12k(3k+1) is divisible by 24.
If k is odd, (3k+1) is even and hence n = 12k(3k+1) is divisible by 24.
If n = 6k+5, (n+1)(n-1) = (6k+6)(6k+4) = 12(k+1)(3k+2).
If k is even, 12(3k+2) is divisible by 24 and hence n = 12(k+1)(3k+2) is divisible by 24.
If k is odd, 12(k+1) is divisible by 24 and hence n = 12(k+1)(3k+2) is divisible by 24.
In either case r = 0.
So, both statements together are sufficient to answer the question.
The correct answer is (C).
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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