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Prove that each natural number has a multiple solely consisting of digits $0,1$
I have really no idea for this!!

Hamid Reza Ebrahimi
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  • every number is divisible by 1... please sharp your question – boaz Jun 07 '16 at 06:43
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    @boaz I don't understand your comment. How do you use the fact that $n$ is divisible by $1$ to prove that some multiple of $n$ has a decimal representation containing only $0$'s and $1$'s? – bof Jun 07 '16 at 06:47
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    Maybe look at the remainders when $1,11,111,1111,11111$, and so on are divided by $m$. Then use the Pigeonhole Principle and it will be nearly all over. – André Nicolas Jun 07 '16 at 06:47
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    The original form of the question was fine; your edit in response to boaz's comment was unnecessary and in fact made the question harder to understand, since $1$ is not a mulitple of most numbers. – joriki Jun 07 '16 at 06:48
  • you right bof...i thought he meant a factor – boaz Jun 07 '16 at 06:54
  • @AndréNicolas But your suggested numbers don't contain digit $0$! That technique looks for a limited part of such multiples – Hamid Reza Ebrahimi Jun 07 '16 at 06:57
  • Hate to give it away entirely. There will be two different strings of $1$'s with the same remainder. Their difference is of the right shape and divisible by $m$. – André Nicolas Jun 07 '16 at 06:59
  • If the number is coprime to $10$, then it is a factor of number of all $1$s. Search our site with the buzzword repunit to find the argument. If your number has $2$s and $5$ as prime factors, zeros will be forced upon you. First multiply the number $n$ by complementary powers of $2$ and $5$ to get something of the form $$2^a5^bn=10^km$$ where $\gcd(m,10)=1$. Added: Mind you, André's argument is even better. – Jyrki Lahtonen Jun 07 '16 at 07:02

2 Answers2

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Consider the remainders when we divide $1,10,10^2,\cdots$ by $k$. We know that at least one of $0,1,\cdots, k-1$ appears infinitely many times. Let it be $r$. Then take $a_i\ (i=1,2,\cdots, k)$ such that $$10^{a_i}\equiv r\pmod k$$ where $0\le a_1\lt a_2\lt \cdots\lt a_k$.

Now

$$10^{a_i}+10^{a_2}+\cdots +10^{a_k}\ \ $$

mathlove
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notice that it is suffices to prove it for prime numbers since the general assertion can be done using induction.

boaz
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