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Let $k \in \mathbb Z^+ $. Prove that there exists a positive integer $n $ such that $k|n$ and the only digits in $n$ are 0's and 3's

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HINT: Let $$m_n=\underbrace{3\dots3}_{n\text{ threes}}\;.$$ Let $r_n$ be the remainder when you divide $m_n$ by $k$. Use the fact that if $r_n=r_\ell$ with $n>\ell$, then $k\mid m_n-m_\ell$.

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the digits in $m_n$ are only 3 or consist on 3 and 0 ? – chihiroasleaf Feb 6 '13 at 7:35
@chihiroasleaf: The digits of $m_n$ are all $3$’s. Note that you’re only required to find a number that has no digit other than $0$ and $3$: it need not have both $0$ and $3$. However, in many cases the number that you get from this hint won’t be one of the $m_n$’s, and it will have both $3$’s and $0$’s. – Brian M. Scott Feb 6 '13 at 7:37
@chihiroasleaf Said differently, consider the set {3, 33, 333, 3333, ...., (3 repeated k times)}. Now, any number when divided by k can give remainders between 0 and k - 1. Now, divide each number in the given set with k and observer the remainders that they give. What happens if one of the remainders is zero and what happens if none of the remainders are zero? – TenaliRaman Feb 6 '13 at 7:46
is this correct? I make $k+1$ $m_n$ they are $m_1$ = 3 $m_2$ = 33 . – chihiroasleaf Feb 6 '13 at 7:46
@chihiroasleaf: Yes, two of $m_1,\dots,m_{k+1}$ must have the same remainder. Now what can you do with them to get a multiple of $k$ that has only $3$’s and maybe some $0$’s? – Brian M. Scott Feb 6 '13 at 7:48

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