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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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

Answer: E

RELATED VIDEO

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its E
pick up the numbers

ex n = 2
2^1/7 = 2
2^2/7 = 4
2^3/7 = 1
2^4/7 = 4
2^5/7 = 4
2^6 /7 = 1

so m can be 3 or 6
now take n = 3
3^3/7 = 6
3^6/7 = 1

hence 6.

concept:
we know that the remainder when an interger is divided by 7 are 1,2,3,4,5,6 (n-1 concept)
now suppose m =2
the remainder will also be raised by m
1,4,9,16,25,36
divide this by 7 we dont get 1 consistently

try for 3,4,5,.. once u get 6 u will get all 1s
1,64,729,4096, 15625 , 46656) all these number when divided by 7 leaves a remainder 1.

hope this helps.

hi everyone,

I solved this question like this , as 7 is a prime no and n can only take values which are not multiples of 7 so n and 7 will be co prime 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

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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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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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

Nice explanation Sudhir.

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

I understand your solution, But how is it associated with Binomial theorem, we have solved it with plugin here?

Can anyone please help to explain this issue! I have the same question!
Another case can be: n=2, m=3: 2^3 = 8 = 7*1 + 1 (n^m = 7*a + 1)

n^m = 7x +1 ; numbers are 8,15,22,29,36,43,50,57,64,..

The squares we see are : 36 = 3^2x2^2 -> cannot be expressed as n^m
Next
64 =2^6 -> expressed in n^m
So 6 is answer. E.

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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Check using least values -

2^2 = 4 ( Will not give remainder 1 when divided by 7 - Eliminate option A )
3^3 = 27 ( Will not give remainder 1 when divided by 7 - Eliminate option B )
2^4 = 16 ( Will not give remainder 1 when divided by 7 - Eliminate option C )
2^5 = 32 ( Will not give remainder 1 when divided by 7 - Eliminate option D )

2^6 = 16 will give remainder 1 when divided by 7
3^6 = 729 2will give remainder 1 when divided by 7

The answer must be (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.

Hi,

we look for integer=m>0 such that n^m e {7a+1}, where a is a positive integer. This is because in this set, we have all the positive integers which leave remainder 1 when divided by 7. In general we know:

(x+y)^z=Mx+(y^z/x), where Mx denotes a multiple of x. To understand why, expand the binomial for any n, and you realize that any term contains an x, except the last one, which will be y^z. We further know that we look for z such that:

(x+y)^z=M7+1 -> (y^z/x)=1 for any x+y, where x and y are positive integers

Now, we have to find a way to depict all positive integers as x+y. If we do so smartly, we choose (7a+1), (7a+2),...,(7a+6). The union of these sets depicts all positive integers. All we are left to do is plug in m and see if (y^z)/7=M7+1