# Sixes and sevens

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.

Note by Bruce Wayne
4 years, 9 months ago

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FOLLOWING IS THE SOLUTION. DON'T READ THIS COMMENT IF YOU HAVEN'T TRIED THE PROBLEM ON YOUR OWN

There 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 $$2^{67}$$.

- 4 years, 9 months ago

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 $$0$$").

- 4 years, 9 months ago

Comment deleted Dec 11, 2013

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.

- 4 years, 9 months ago