Prove that any positive integer has a multiple whose decimal expansion involves all ten digits.

Question

This is something I've been struggling with - I've been having trouble even wrapping my head around what is being asked. It feels like the answer might be on or around something extremely obvious, but my failure to understand the statement is keeping me from seeing it. Even just a broader or more detailed explanation of the question being asked, if someone else can see what it is, would be beyond helpful.

Answer 1

To prove something a bit stronger:

Claim: Given two fixed natural numbers, $N,M$, there is a multiple of $N$ the decimal expansion of which contains $M$ as a string of digits (With $N=267, M=711$ we see $267\times 139=37113$ as an example).

Proof: Consider the sequence which concatenates $M$ with itself. Thus, if $L(M)$ denotes the number of digits in $M$, we have $$a_1=M\quad a_n=a_{n-1}\times 10^{L(M)}+M$$.

Reduce this sequence $\pmod N$. As there are only finitely many residues $\pmod N$ we must be able to find $i<j$ with $a_i\equiv a_j\pmod N$ and then $(a_j-a_i)$ satisfies the claim.

Note: Inspecting the proof shows that we can find a multiple of $N$ which includes only the digits of $M$ and $0$.

To answer the question at hand, take $M=1234567890$ or any other natural number that includes all the digits.

Answer 2

Prove instead: Every integer n >= 1 has a multiple whose decimal representation starts with 1234567890.

And that’s simple: If your n has k digits, then take 1234567890, followed by k zeroes, divide by n and round up to m. m times n is at least 1234567890 with k zeroes, but the rounding up cannot change the first 10 digits.