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let M be a set of those natural numbers that can be written using only 0's and 1's(in the decimal system).prove that for evey natural number k,there exist a number m ∈ M such that (i) m has exactly k 1's,and (ii) m is divisible by k. I know bits and pieces of the proofs but it just not looking right to me...please help

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    http://math.stackexchange.com/questions/2118246/prove-that-for-any-natural-number-k-exists-m-in-m-s-t-i-m-has-exactly – 2017-02-23

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Hint. The sequence $(10^n \mod k)_{n \geq 0}$ is eventually periodic.

Thus you just pick $k$ different instances from this sequence with the same value and this gives rise to the desired $m \in M$.

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    this actually gave me a direction,so do I have to pick different values of k and do they have to be in an increasing manner in a way that the greater the k is the more it will secure the proof? for example if m=101010 then k=3 right? – 2017-02-17
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    $k$ is some fixed number, you cannot pick it. – 2017-02-17