Can the last digit in the base ten expansion of X^2X be 4, whenever X is a positive integer?

If so, give an example. If not, prove it.

If X is odd, then X^2X is odd and cannot end in 4. If X ends in 0, then X^2X ends in 0. Suppose X ends in 2, 4, 6, or 8. Since X is even, X=2Y for some integer Y. Then, X^2X=X^4Y=(X^4)^Y. Since X ends in 2, 4, 6, or 8, X^4 ends in 6. Since every power of a number ending in 6 ends in 6, X^2X=(X^4)^Y ends in 6. Therefore, X^2X can never end in 4.

Posted by Math Man on 2014-03-19 21:04:12