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Re: OG DS - Don't understand the answer [#permalink]
21 Jan 2007, 01:57

focused07 wrote:

If n is a positive integer, is n^3 - n divisible by 4?

(1) n = 2k + 1, where k is an integer. (2) n^2 + n is divisible by 6.

OG has a convoluted explanation that gives (A) as the answer.

However I solve it using plug-in method and get (E). Could someone please tell me where I made a mistake.

Statement 1

n = 2 (K) + 1 , where K is an integer.

As per statement 1, If K is an integer, then K can be -2 , -1, 0 , 1, 2, 3

When K = 0 ( based on problem statement, K is an integer ) n = 2 ( 0) + 1 = 1

When K = - 1 ( based on problem statement, K is an integer ) n = 2 ( - 1) + 1 = - 1

When k = 1 n = 2 ( 1) + 1 = 3

Taking values of K where n is a positive integer

If K = 0 , then n = 1 , and n^3 â€“ n = 0

Therefore for statement 1 when K is the 0, the statement is insufficient for n^3 â€“ n to be divisible by 4.

Statement 1 is sufficient only when k is a positive integer greater than 0 , a condition not explicitly given in statement 1.

Statement 2

Insufficient.

Both Statement 1 and 2 are insufficient when taken together

Therefore (E)

In bold, I show the problems in your reasoning

In the first bold, k cannot be negative because n is a positive integer, definied such as this.

In the second bold, 0/4 = 0. 0 is disivable by any integer. Why? For any integer a, we have 0/a = 0*k + 0. The remainder is always equal to 0, the definition of divisibility

Last edited by Fig on 21 Jan 2007, 03:40, edited 3 times in total.

To do DS divisibility and remainder problems, it really helps to understand how to write and interpret things written in mod form, where:

n = dk + r, where n is an integer, d is an integer divisor, k is any integer, and r is an integer remainder

Examples:

x is divisible by 3 can be written: x = 3k + 0, for any integer k
x, when divided by 3 has a remainder of 1: x = 3k + 1, for any integer k

In this problem, statement one states:

n = 2k + 1, for any integer k. What this means is that when n is divided by 2, there is a remainder of 1. This in turn is just a convoluted way of saying n is an odd integer.

If you plug any odd integars into the formual n^3 - n, you'll find that they are divisible by four.