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Re: From GMAT Prep [#permalink]
11 Jan 2009, 17:25

sid3699 wrote:

Could someone please explain the answer. Thanks.

If n is a postive integer and r is the remainder when (n+1)(n-1) is divided by 24, what is the value of r?

(1) n is not divisible by 2

(2) n is not divisible by 3

Edited by Gmat Tiger (GT): No OA with the question; post OA only after 4/5 responses.

individually n is not divisible by 2=> n belongs to 3, 5,7,9,11,13,15,17,19,................ (odd numbers) n^2-1:8,24,48,80. so r can be 0 or non-zero. hence not sufficient. (2) n belongs to the series: 2,4,5,7,8,10,11,13,........... n^2-1:3,15,24,48,63, again remainder is not determinate. hence not sufficient. (1) & (2) together define the following series: 5,7,11,13,17,19,23,25,29,31,35,37,.............. n^2-1:24,48,120,168,....... hence sufficient. Hence, answer C. I have perhaps proved this whole thing; but anyone has an easier way?

Re: From GMAT Prep [#permalink]
12 Jan 2009, 01:36

sid3699 wrote:

Could someone please explain the answer. Thanks.

If n is a postive integer and r is the remainder when (n+1)(n-1) is divided by 24, what is the value of r?

(1) n is not divisible by 2

(2) n is not divisible by 3

Edited by Gmat Tiger (GT): No OA with the question; post OA only after 4/5 responses.

There is an easier way, yes, if you know a few basic properties from number theory.

First, if you take three consecutive integers:

n-1, n, n+1

it will always be true that exactly one of these integers is divisible by 3 (since multiples of 3 are separated by 3 on the number line).

Second, say n-1 and n+1 are two consecutive even numbers. Then it must be true that one of them is a multiple of 4, since every second even number is a multiple of 4.

Now to the question: n-1, n, n+1 are three consecutive integers. If we use both statements, we know that n is not a multiple of 3 , so either n-1 or n+1 must be a multiple of 3. We also know that n is odd, so n-1 and n+1 are both even, and one of them is a multiple of 4. So the product (n-1)(n+1) must be divisible by 3*2*4 = 24, and if (n-1)(n+1) is a multiple of 24, the remainder will be zero when we divide it by 24.

Since neither statement is sufficient alone, the answer is C.

_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: From GMAT Prep [#permalink]
12 Jan 2009, 05:40

great man. I always tend to go back to deriving it.

Do you happen to have any website\notes\links where such properties are summarized?

IanStewart wrote:

sid3699 wrote:

Could someone please explain the answer. Thanks.

If n is a postive integer and r is the remainder when (n+1)(n-1) is divided by 24, what is the value of r?

(1) n is not divisible by 2

(2) n is not divisible by 3

Edited by Gmat Tiger (GT): No OA with the question; post OA only after 4/5 responses.

There is an easier way, yes, if you know a few basic properties from number theory.

First, if you take three consecutive integers:

n-1, n, n+1

it will always be true that exactly one of these integers is divisible by 3 (since multiples of 3 are separated by 3 on the number line).

Second, say n-1 and n+1 are two consecutive even numbers. Then it must be true that one of them is a multiple of 4, since every second even number is a multiple of 4.

Now to the question: n-1, n, n+1 are three consecutive integers. If we use both statements, we know that n is not a multiple of 3 , so either n-1 or n+1 must be a multiple of 3. We also know that n is odd, so n-1 and n+1 are both even, and one of them is a multiple of 4. So the product (n-1)(n+1) must be divisible by 3*2*4 = 24, and if (n-1)(n+1) is a multiple of 24, the remainder will be zero when we divide it by 24.

Since neither statement is sufficient alone, the answer is C.

_________________

----------------------- tusharvk

gmatclubot

Re: From GMAT Prep
[#permalink]
12 Jan 2009, 05:40