How many (Base 10) 67-digit multiples of \(2^{67}\) exist which can be written exclusively with the digits 6 and 7 ?

**SOLUTION HAS BEEN ADDED AS A COMMENT**

As an additional exercise, try the following:

Show that every number which is not divisible by 5 has a multiple whose only digits are 6 and 7.

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## Comments

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TopNewestFOLLOWING IS THE SOLUTION. DON'T READ THIS COMMENT IF YOU HAVEN'T TRIED THE PROBLEM ON YOUR OWNThere are exactly \(2^{67}\) 67-digit numbers containing only the digits 6 and 7. The difference of any two of these is not divisible by \(2^{67}\) since it is expressible as the sum or difference of powers of 10.

Hence, these \(2^{67}\) numbers have pairwise different remainders on division by \(2^{67}\), and since there are \(2^{67}\) such remainders (from \(0\) to \(2^{67}-1\))there must be (exactly one) which has remainder 0 and hence is divisible by \(2^{67}\).

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I don't think the fact that the difference being expressible as the sum or difference of powers of \(10\) trivially concludes that it's not divisible by \(2^{67}\), although I've seen a similar problem before which elaborates further (somewhere along "\(\displaystyle\sum_{i=0}^{k-1} a_i10^i\) where \(a_i \in \{-1,0,1\}\) is not divisible by \(2^k\) except when it's equal to \(0\)").

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Comment deleted Dec 11, 2013

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Please avoid adding answers which do not increase the knowledge of the community. This is an actual question that I came across. I was amazed on seeing the simple elegant solution to this problem and thus wanted to share it with everyone. If you have a solution, please share it.

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