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

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

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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!

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

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?

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.

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)

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.

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

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)

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

Given that n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7 and we need to find m could be equal to

Let's take each option choice and check which one satisfies the statement

(A) 2
=> \(n^2\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^2\) = 4 divided by 7 gives 4 remainder ≠ 1 => NOT POSSIBLE

(B) 3
=> \(n^3\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^3\) = 8 divided by 7 gives 1 remainder
Let's take n=3 we get
\(3^3\) = 27 divided by 7 gives 6 remainder ≠ 1 => NOT POSSIBLE

(C) 4
=> \(n^4\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^4\) = 16 divided by 7 gives 2 remainder ≠ 1 => NOT POSSIBLE

(D) 5
=> \(n^5\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^5\) = 32 divided by 7 gives 4 remainder ≠ 1 => NOT POSSIBLE

(E) 6
So, Answer will be E but let's check for 1 number
=> \(n^6\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7
Let's take n=2 we get
\(2^6\) = 64 divided by 7 gives 1 remainder => POSSIBLE

So, Answer will be E
Hope it helps!

Watch the following video to learn the Basics of Remainders