1998 RMO

Suppose that a number $n$ has final digit $b$ . Then $n = 10a+b$ for some integer $a$ . After rubbing out the unit digit $b$ and adding $5b$ to what remains, we have $\hat{n} = a+5b$ . Since $\hat{n} + 2n = (a+5b) + 2(10a + b) = 7(3a + b)$ , if $n$ is a multiple of $7$ , then so is $\hat{n}$ . If we start with $7^{1998}$ , any subsequent number obtained by this process must be a multiple of $7$ , and $1998^7$ is not a multiple of $7$ . Thus we can never end up with $1998^7$ .