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.