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If n^m leaves a remainder of 1 after division by 7 for all

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If n^m leaves a remainder of 1 after division by 7 for all

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If \(n^m\) leaves a remainder of 1 after division by 7 for all positive integers \(n\) that are not multiples of 7, then \(m\) could be equal to:

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

OA=E

Source: Manhattan GMAT

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Gmat_mission wrote:
If \(n^m\) leaves a remainder of 1 after division by 7 for all positive integers \(n\) that are not multiples of 7, then \(m\) could be equal to:

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

OA=E

Source: Manhattan GMAT
The most important piece of information is here: "for all positive integers n that are not multiples of 7"

Since 2 is not a multiple of 7, then it must be the case that, for a particular value of m, 2^m leaves a remainder of 1 after division by 7

Let's check the answer choices....
(A) if m = 2, we get 2^2 = 4.
When we divide 4 by 7, we get a remainder of 4. We need a remainder of 1. ELIMINATE A

(B) if m = 3, we get 2^3 = 8.
When we divide 8 by 7, we get a remainder of 1. KEEP B

(C) if m = 4, we get 2^4 = 16.
When we divide 16 by 7, we get a remainder of 2. We need a remainder of 1. ELIMINATE C

(D) if m = 5, we get 2^5 = 32.
When we divide 32 by 7, we get a remainder of 4. We need a remainder of 1. ELIMINATE D

(E) if m = 6, we get 2^6 = 64.
When we divide 64 by 7, we get a remainder of 1. KEEP E

So, the correct answer is either B or E

Now try a different value of n.
How about n = 3
Check the remaining answer choices....

(B) if m = 3, we get 3^3 = 27.
When we divide 27 by 7, we get a remainder of 6. ELIMINATE B

_________________
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Gmat_mission wrote:
If \(n^m\) leaves a remainder of 1 after division by 7 for all positive integers \(n\) that are not multiples of 7, then \(m\) could be equal to:

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

OA=E

Source: Manhattan GMAT
We have to analyze each given answer choice.

A) 2

We see that 1^2 = 1 and 2^2 = 4. While the former leaves a remainder of 1 when it’s divided by 7, the latter leaves a remainder of 4 when it’s divided by 7. So m couldn’t be 2.

B) 3

We see that 1^3 = 1, 2^3 = 8 and 3^3 = 27. While the first two leave a remainder of 1 when they are divided by 7, the last one leaves a remainder of 6 when it’s divided by 7. So m couldn’t be 3.

C) 4

We see that 1^4 = 1 and 2^4 = 16. While the former leaves a remainder of 1 when it’s divided by 7, the latter leaves a remainder of 2 when it’s divided by 7. So m couldn’t be 4.

D) 5

We see that 1^5 = 1 and 2^5 = 32. While the former leaves a remainder of 1 when it’s divided by 7, the latter leaves a remainder of 4 when it’s divided by 7. So m couldn’t be 5.

So E must be the correct answer, but let’s verify it, anyway.

E) 6

We see that 1^6 = 1, 2^6 = 64, and 3^6 = 729. We see that all these numbers leave a remainder of 1 when they are divided by 7. So m could be 6.

Answer: E

_________________

Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



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