Let u(n) be 11...1 (n times)

Let x be a natural odd number undividable by number 5. This implies x is relatively prime to any multiple of 10 because their divisors are only 5 and 2.

Using the Pigeonhole principle - Wikipedia, the free encyclopedia there is p and q such that the remainders of x/u(p) and x/u(q) are the same so x divide a number written 11...100..0 (1's followed by 0's)

x divide 11...1 * 10...0 and x is relatively prime to 10...0 so x must divide 11...1 using this lemma : If n|ab, and n is relatively prime to a, then n|b.