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

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

The OA is E.

I solved this PS question like this, as 7 is a prime number and n can only take values which are not multiples of 7 so, n and 7 will be coprime hence as per Fermat little theorem value of m will be 7 - 1 = 6 for any value of n which are not multiples of n.

Can someone explain another way to solve this PS question? Thanks!

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

by Vincen » Thu Jun 28, 2018 2:39 am

Hello BTGmoderatorLU.

I would solve it as follows.

Since the statement must be true for all values of n that are not multiples of 7, then I picked n=3.

Now, let's try the values of m.

If m=2 then 3^2 = 9, remainder 2. Out.

If m=3 then 3^3 = 27, remainder 6. Out.

If m=4 then 3^4 = 81, remainder 4. Out.

If m=5 then 3^5 = 243, remainder 5. Out.

If m=6 then 3^6 = 729, remainder 1. Correct.

So, the correct answer is the option E.

I hope it helps you. <i class="em em-smiley"></i>

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Mon Jul 02, 2018 9:07 am

BTGmoderatorLU 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

Let's test each answer choice.

A. m = 2

If n = 2, we see that 2^2/7 = 4/7 = 0 R 4, so m could not be 2.

B. m = 3

If n = 3, we see that 3^3/7 = 27/7 = 3 R 6, so m could not be 3.

C. m = 4

If n = 2, we see that 2^4/7 = 16/7 = 2 R 2, so m could not be 4.

D. m = 5

If n = 2, we see that 2^5/7 = 32/7 = 4 R 4, so m could not be 5.

At this point, we see that the correct choice must be E, but let's verify that is case anyway.

E. m = 6

If n = 1, we see that 1^6/7 = 1/7 = 0 R 1.
If n = 2, we see that 2^6/7 = 64/7 = 9 R 1.
If n = 3, we see that 3^6/7 = 729/7 = 104 R 1.

We can stop at this point before the numbers get too large; we can see that E is the correct answer choice.

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

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Shahrukh_mbabreakspace: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Mon Jul 02, 2018 1:56 am; Followed by:2 members

by Shahrukh_mbabreakspace » Tue Jul 03, 2018 4:00 am

Fermat little theorem is the best way to solve it.
We can also do it using Euler's function as well
7*(1-1/7)= 6

________________________________________

Shahrukh Moin Khan
QA Mentor at Breakspace
https://www.mbabreakspace.com
PGDM: IIM Calcutta
B.Tech IIT Roorkee

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