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N is a positive integer. 36^N and 37^N are divided by 7 with

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SVP
SVP
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Joined: 03 Feb 2003
Posts: 1611
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Kudos [?]: 64 [0], given: 0

N is a positive integer. 36^N and 37^N are divided by 7 with [#permalink] New post 12 Oct 2003, 23:43
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N is a positive integer. 36^N and 37^N are divided by 7 with some remainders. Find the smallest N when the remainders are the same. Find the next N.
CEO
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Joined: 15 Aug 2003
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Kudos [?]: 684 [0], given: 781

Re: PS: TRIBUTE TO THE SAME REMAINDER [#permalink] New post 13 Oct 2003, 00:48
stolyar wrote:
N is a positive integer. 36^N and 37^N are divided by 7 with some remainders. Find the smallest N when the remainders are the same. Find the next N.



n=1

36^ N divided by 7 leaves remainder of 1

37^N divided by 7 leaves remainder of 2

n =2

36^ 2 divided by 7 leaves remainder of 1^2 =1

37^2 divided by 7 leaves remainder of 2^2 =4

n= 3

36^3 divided by 7 leaves remainder of 1^3 = 1
37^2 divided by 7 leaves remainder of 2^3 = 8

since 8>7 , divide 8/7 , we get remainder of 1 ...

N = 3 is the answer....


Next N

Since we already discussed a variant of this question...allow me to do this

faster.

The next highest cube of 2 that when divided by 7 leaves a remainder 1 is

2^6 = 64 ... 64/7 leaves remainder of 1.

so , Next N =6

thanks
praetorian
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 [#permalink] New post 13 Oct 2003, 01:01
agree
a very interesting math problem! where did you get an idea?
  [#permalink] 13 Oct 2003, 01:01
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N is a positive integer. 36^N and 37^N are divided by 7 with

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