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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 _________________

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.

Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]
09 Aug 2013, 04:07

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! _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]
09 Aug 2013, 04:13

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

Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]
08 Sep 2013, 07:55

abhishekkhosla wrote:

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

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. _________________

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.

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

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.

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. _________________