Got one from GMATPrep

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Got one from GMATPrep

by netneoaaa » Thu Jul 05, 2007 3:50 am
n is an integer >0 r is remainder when n^2-1 div by 24. What is r ?

1. n not div by 2
2. n not div by 3

OA = both (c)

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n^2-1=24t+r
r=n^2-1-24t

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n<>2k
n<>3k how to proceed

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by givemeanid » Thu Jul 05, 2007 6:27 am
(1) n mod 2 != 0
n = 3, n^2-1 = 8. Remainder = 8
n = 5, n^2-1 = 24. Remainder = 0
Not sufficient.

(2) n mod 3 != 0
n = 2, n^2-1 = 3. Remainder = 3
n = 5, n^2-1 = 24. Remainder = 0
Not sufficient.


Now, putting both together:
n mod 2 != 0
n mod 3 != 0

n^2-1 = (n+1)(n-1)
When n mod 2 != 0, n is odd. So, both n+1 and n-1 are divisible by 2. Hence, (n+1)(n-1) is divisible by 4.
When n mod 3 != 0, either n-1 or n+1 is divisible by 3. (n-1, n and n+1 are 3 consecutive integers).

Also, since n is odd AND not divisible by 3, either n-1 or n+1 is divisible by 4. Now, we accounted for either n-1 or n+1 divisible by 2 using the statement (1). So, (n-1)(n+1) is divisible by 8.

Combining, (n-1)(n+1) is divisible by 8*3 = 24.
Hence, remainder when n^2-1 is divided by 24 is 0.
r=0.


Answer is (C).
So It Goes

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by bingojohn » Wed Aug 08, 2007 9:23 am
givemeanid wrote:(1) n mod 2 != 0
n = 3, n^2-1 = 8. Remainder = 8
n = 5, n^2-1 = 24. Remainder = 0
Not sufficient.

(2) n mod 3 != 0
n = 2, n^2-1 = 3. Remainder = 3
n = 5, n^2-1 = 24. Remainder = 0
Not sufficient.


Now, putting both together:
n mod 2 != 0
n mod 3 != 0

n^2-1 = (n+1)(n-1)
When n mod 2 != 0, n is odd. So, both n+1 and n-1 are divisible by 2. Hence, (n+1)(n-1) is divisible by 4.
When n mod 3 != 0, either n-1 or n+1 is divisible by 3. (n-1, n and n+1 are 3 consecutive integers).

Also, since n is odd AND not divisible by 3, either n-1 or n+1 is divisible by 4. Now, we accounted for either n-1 or n+1 divisible by 2 using the statement (1). So, (n-1)(n+1) is divisible by 8.

Combining, (n-1)(n+1) is divisible by 8*3 = 24.
Hence, remainder when n^2-1 is divided by 24 is 0.
r=0.


Answer is (C).
Nice problem... awesome solution... great thinking