Responding to a pm:

It is a conceptual question and it easy to figure out if you understand binomial theorem discussed here: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

n^m leaves a remainder of 1 after division by 7 for all n (obviously n cannot be a multiple of 7 because that would leave a remainder of 0)
So n can be of the form (7a + 1) or (7a + 2) or (7a + 3) or (7a + 4) or (7a +5) or (7a + 6)

According to binomial, the remainder I will get when I divide n^m by 7 will depend on the last term i.e. 1^m or 2^m or 3^m or 4^m or 5^m or 6^m.
We need a value of m such that when we divide any one of these 6 terms by 7, we always get a remainder 1.

Can m be 2? 1^2 leaves remainder 1 but 2^2 leaves remainder 4. So no
Can m be 3? 1^3 leaves remainder 1, 2^3 leaves remainder 1. 3^3 leaves remainder 6. So no
Can m be 4? 1^4 leaves remainder 1 but 2^4 leaves remainder 2. So no
Can m be 5? 1^5 leaves remainder 1 but 2^5 leaves remainder 4. So no'

m must be 6 because that is the only option left.
1^6 leaves remainder 1, 2^6 leaves remainder 1 etc. So we see that we are correct.
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We need to find m for all positive integers.
Therefore, if we take a case of 2 then \(2^3\) and \(2^6\) will leave remainder of 1 after dividing by 7. we will restrict to 6 and not more than that because it is a largest value in the options and we have to select only one of them.

Then, i took a case of 3 ...In that 3^6 leaves remainder of 1 when divided by 7.

Now, we can conclude that the value of m will be 6 because it common value in both cases and one of the given option has to be correct. Hence m=6
thus E

Somehow didn't get this question as per below.

n^m = 7Q + 1

n != 7A (where != is not equal to)

Consider n=6 and answer option (A) m=2

6^2 = 36 = 7*5 + 1

Why not (A) then?

Rgds,
TGC!

for those who know fermat little theorem(s) its good but for others, plugging in will always help as explained by Sudhir... thanks Abhishek though for reminding this theorem

The remainder you get when you divide (7a + 1)^m by 7 will be 1. The remainder you get when you divide (7a + 2)^m by 7 is determined by 2^m. This is determined by binomial theorem. The link explains you why.
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The question says that remainder should be 1 for all values of n. So n could be 1 or 2 or 3 or 4 etc, remainder when n^m is divided by 7 will ALWAYS be 1. Check for a few values of n.

In case m = 2,
1^2 = 1 - when 1 is divided by 7, remainder is 1 - fine
2^2 = 4 - when 4 is divided by 7, remainder is 4 - not acceptable

In case m = 3,
1^3 = 1 - when 1 is divided by 7, remainder is 1 - fine
2^3 = 8 - when 8 is divided by 7, remainder is 1 - fine
3^3 = 27 - when 27 is divided by 7, remainder is 6 - not acceptable

Only in case m = 6, for every value of n, you will get remainder 1.

Answer (E)

Another thing, if m = 2 or m = 3 were the answer, m = 6 would automatically be the answer too because

n^6 = (n^3)^2 = (n^2)^3

But in problem solving questions, you have only one correct answer.
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Option E satisfies the required condition (remainder is 1 when n^m is divided by 7) for all values of 'n' whereas option B satisfies it only when n=2 (remainder is 1 when 2^3 is divided by 7 but it is 6 when n=3: 3^3=27/7). Actually, the question itself is confusing. It should have stipulated: "m must be (instead of, could be) equal to:".

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
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The learning from this question is to test each case for a "could be " question. I stopped at B once I determined that 2^3 worked.