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by adthedaddy » Mon Aug 20, 2012 6:40 am
Case A: n is a prime number
If n=2, (2)^2-1= 4-1= 3 => Not Divisible by 24
If n=5, (5)^2-1= 24 => Divisible by 24.
Thus we do not get a unique value. Therefore, NOT SUFFICIENT

Case B: n>191
Not sufficient as we get different answers for n=24^2 and n=17^2

Combined:
n is prime and greater than 191.
Thus, n is odd and not a multiple of 3.
As n^2-1 = (n-1)(n+1), either of (n-1),(n+1) is a multiple of 3.
Also, as n is odd, (n-1) and (n+1) are even.
So, one of them must be a multiple of 4. Thus, (n-1)(n+1)=2*4=8.
Thus, for a number to be divisible by 24, it should be divisible by 8 & 3.

Thus answer is (C).
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by Anurag@Gurome » Mon Aug 20, 2012 8:43 pm
neer.king wrote:If n is a positive integer, is n^2-1 divisible by 24?

A: n is a prime number
B: n is greater than 191
(1) n is a prime number.
If n = 2, then n² - 1 = 4 - 1 = 3, which is not divisible by 24.
If n = 5, then n² - 1 = 25 - 1 = 24, which is not divisible by 24.
No definite answer; NOT sufficient.

(2) n is greater than 191. Clearly insufficient (consider n=24^2 for a NO answer and n=17^2 for an YES answer).
If n = 17² = 289, then n² - 1 = (n - 1)(n + 1) = 288 * 290, which is divisible by 24.
If n = 24² = 576, then n² - 1 = (n - 1)(n + 1) = 575 * 577, which is NOT divisible by 24.
No definite answer; NOT sufficient.

Combining (1) and (2), n is a prime number so that n > 191.
This means n is odd and not a multiple of 3.
n² - 1 = (n - 1)(n + 1) implies out of three consecutive integers (n - 1), n and n + 1, one must be divisible by 3, since it's not n then it must be either (n - 1) or (n + 1), so (n - 1)(n + 1) is divisible by 3.
Since n is odd then (n - 1) and (n + 1) are consecutive even numbers, which means that one of them must be a multiple of 4, so (n - 1)(n + 1) is divisible by 2 * 4 = 8.
We have (n - 1)(n + 1) is divisible by both 3 and 8 so (n - 1)(n + 1) is divisible by 3 * 8 = 24; SUFFICIENT.

The correct answer is C.
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