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
[spoiler]OA=E[/spoiler]
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
Brent Hanneson - Creator of GMATPrepNow.com
