Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.