<pre>
Let n be our integer; one such desired multiple is then
(10^(phi(n))-1)/9.  All we need is that (n,10) = 1, and if the last
digit is 3 this must be the case.  A different proof using the
pigeonhole principle is to consider the sequence 1, 11, 111, ...,
(10^n - 1)/9.  We must have at some point that either some member of our
sequence = 0 (mod n) or else some value (mod n) is duplicated.  Assume
the latter, with x''a and x''b, x''b>x''a,  possesing the duplicated
remainders.  We then have that x''b - x''a = 0 (mod n).  Let m be the
highest power of 10 dividing x''b - x''a.  Now since (10,n) = 1, we can
divide by 10^m and get that (x''b - x''a)/10^m = 0 (n).  But (x_b -
x_a)/10^m is a number containing only the digit 1.

Q.E.D.
</pre>
