Yes
No

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

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 $1998^7$ .