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If m and n are positive integers, is n^m - n divisible by 6?

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If m and n are positive integers, is n^m - n divisible by 6?  [#permalink]

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New post 28 Nov 2016, 01:43
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If m and n are positive integers, is \(n^m - n\) divisible by 6?

(1) m = 3
(2) n = 2

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Re: If m and n are positive integers, is n^m - n divisible by 6?  [#permalink]

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New post 28 Nov 2016, 02:13
MathRevolution wrote:
If m and n are positive integers, is \(n^m\)-n divisible by 6?

1) m=3
2) n=2


This is an interesting sum in my opinion.

if we now the format directly , we can choose the answer.


For Statement 1: m=3 , say n=2 , so 8-2 / 6 --- it works
say n=3 , so 27-3 / 6 ---- it works

so A
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Re: If m and n are positive integers, is n^m - n divisible by 6?  [#permalink]

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New post 28 Nov 2016, 02:20
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MathRevolution wrote:
If m and n are positive integers, is \(n^m\)-n divisible by 6?

1) m=3
2) n=2


Statement 1
m = 3 therefore, is n^3-n divisible by 6?
Using n = 1, 1^3-1 = 0 which is divisible by 6.
Using n = 2, 2^3-2 = 6 which is divisible by 6.
Using n = 3, 3^3-3 = 24 which is divisible by 6.
Using n = 4, 4^3-4 = 60 which is divisible by 6.
Using n = 7, 7^3-7 = 336 which is divisible by 6.
Using n = 10, 10^3-10 = 990 which is divisible by 6.

Statement 1 is sufficient

Statement 2
n = 2, therefore is 2^m-2 divisible by 6?

using m = 1, 2^1-2 = 0 which is divisible by 6
using m = 2, 2^2-2 = 4 which is not divisible by 6
using m = 3, 2^3-2 = 6 which is divisible by 6
using m = 4, 2^4-2 = 14 which is not divisible by 6
using m = 5, 2^5-2 = 30 which is divisible by 6

Statement 2 is not sufficient.

Answer is A.
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Re: If m and n are positive integers, is n^m - n divisible by 6?  [#permalink]

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New post 28 Nov 2016, 09:14
MathRevolution wrote:
If m and n are positive integers, is \(n^m\)-n divisible by 6?

1) m=3
2) n=2



(n^m)-n= n{[n^(m-1)]-1)

1. so if m=3 then n{(n^2)-1)=(n-1)(n)(n+1)......as this is 3 consecutive integers, it is always divisible by 6

2. if n=2 then try for m=2, 2^2-2=2......do not satisfy
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Re: If m and n are positive integers, is n^m - n divisible by 6?  [#permalink]

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New post 28 Nov 2016, 09:41
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MathRevolution wrote:
If m and n are positive integers, is \(n^m\) - n divisible by 6?

1) m = 3
2) n = 2


Nice question!

Target question: Is \(n^m\) - n divisible by 6?

Statement 1: m = 3
So, we need to determine whether n³ - n is divisible by 6
Factor to get: n³ - n = n(n² - 1)
= n(n + 1)(n - 1)

IMPORTANT: Notice that n-1, n, and n+1 are 3 consecutive integers.
There's a nice rule says: The product of any k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
Likewise, the product of any 11 consecutive integers will be divisible by 11, 10, 9, . . . 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.

This means that the product n(n + 1)(n - 1) [aka n³ - n] is divisible by 3 AND 2, which means it is also divisible by 6.
So, we can be certain that n^m - n is divisible by 6.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: n = 2
Consider the two conflicting cases:
Case a: m = 5 and n = 2, in which case, n^m - n = 2^5 - 2 = 32 - 2 = 30, which IS divisible by 6. In this case, n^m - n IS divisible by 6
Case b: m = 4 and n = 2, in which case, n^m - n = 2^4 - 2 = 16 - 2 = 14, which is NOT divisible by 6. In this case, n^m - n is NOT divisible by 6
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer:

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Re: If m and n are positive integers, is n^m - n divisible by 6?  [#permalink]

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New post 30 Nov 2016, 03:08
==> In the original condition, there are 2 variables (m,n), and in order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer. By solving con 1) and con 2), from\(2^3-2=6\), you get yes, and hence it is sufficient. The answer is C. However, this question is an integer question, one of the key questions, so you need to apply CMT 4. For con 1), from \(n^3-n=(n-1)n(n+1)\), it is the multiple of the three consecutive integers, which always becomes the multiple of 6, hence yes, it is sufficient. For con 2), from n=2 and m=3 yes, m=2 no, and hence it is not sufficient. Therefore, the answer is A.

Answer: A
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Re: If m and n are positive integers, is n^m - n divisible by 6?  [#permalink]

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New post 30 Nov 2016, 12:46
MathRevolution wrote:
If m and n are positive integers, is \(n^m\)-n divisible by 6?

1) m=3
2) n=2


Check the options by pluggin in some values


FROM STATEMENT - I ( SUFFICIENT )

Let n = 2 ; \(n^m\)-n = \(2^3\) - 2 = 6 divisible by 6
Let n = 3 ; \(n^m\)-n = \(3^3\) - 3 = 24 divisible by 6
Let n = 4 ; \(n^m\)-n = \(4^3\) - 4 = 60 divisible by 6
Let n = 5 ; \(n^m\)-n = \(5^3\) - 5 = 120 divisible by 6


Thus, this statement is sufficient...

FROM STATEMENT - II ( INSUFFICIENT )

Let m = 2 ; \(n^m\) - n = \(2^2\) - 2 = 2 not divisible by 6
Let m = 3 ; \(n^m\) - n = \(2^3\) - 2 = 6 divisible by 6

Thus, this statement can not give us a unique solution...

Hence, Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked, answer will be (A)


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Re: If m and n are positive integers, is n^m - n divisible by 6?  [#permalink]

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Re: If m and n are positive integers, is n^m - n divisible by 6?   [#permalink] 16 Aug 2019, 01:59
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