Spoiler Alert! The number in general (I said, Spoiler Alert!) can be written as _y_ _x_ where 0 <= x <= 9 and the value of the number is 10y + x. Then it must

Spoiler Alert! The number in general (I said, Spoiler Alert!) can be written as _y_ _x_ where 0 <= x <= 9 and the value of the number is 10y + x. Then it must satisfy for some k 2(10y+x) = (10^k)x + y Simplify that and you can show that 10^k - 2 must be divisible by 19, which means k=17 (or 35, or 53, ...). Once you have that, pick x = 2, 3, or 4 and solve for y, and you have your 18-digit number. (Letting x = 1 gives you an 18-digit numbers whose first digit is 0, which doesn't count.)