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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.