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

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N is a positive integer. 36^N and 37^N are divided by 7 with [#permalink]

13 Oct 2003, 00:43

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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.

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Re: PS: TRIBUTE TO THE SAME REMAINDER [#permalink]

13 Oct 2003, 01:48

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]

13 Oct 2003, 02:01

agree
a very interesting math problem! where did you get an idea?

[#permalink] 13 Oct 2003, 02:01

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

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

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

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