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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. _________________
Karishma GMAT Instructor at Angles and Arguments https://anglesandarguments.com/
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.
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
Re: If n^m leaves a remainder of 1 after division by 7 for all p
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09 Aug 2013, 04:13
2
Kudos
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
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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.
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. _________________
Karishma GMAT Instructor at Angles and Arguments https://anglesandarguments.com/
Re: If n^m leaves a remainder of 1 after division by 7 for all p
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14 Jul 2015, 00:41
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Expert Reply
Beat720 wrote:
TGC wrote:
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!
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. _________________
Karishma GMAT Instructor at Angles and Arguments https://anglesandarguments.com/
Re: If n^m leaves a remainder of 1 after division by 7 for all p
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29 Aug 2017, 23:05
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:".
Re: If n^m leaves a remainder of 1 after division by 7 for all p
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02 Jul 2018, 10:23
AnkitK 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
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)
_________________
Thanks and Regards
Abhishek....
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Re: If n^m leaves a remainder of 1 after division by 7 for all p
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07 Jan 2022, 02:59
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
gmatclubot
Re: If n^m leaves a remainder of 1 after division by 7 for all p [#permalink]
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