Question

$19^{19} + 99^{99}$ 的个位数字是多少？

What is the units digit of $19^{19} + 99^{99}?$

$0$

$1$

$2$

$8$

$9$

幂的个位数字只取决于底数的个位数字。

观察可知， $9$ 的偶次幂个位是 $1$ ，奇次幂个位是 $9$ 。

因此 $19^{19}$ 个位是 $9$ ， $99^{99}$ 个位也是 $9$ 。相加后的个位为 $8$ 。

所以正确答案是 D。

Note that the units digit of an exponent depends only upon the units digit of the base.

Experimenting, we get that $9$ to even power ends with a $1$ and to an odd power ends with a $9.$

Therefore, $19^{19}$ ends with a $9$ and $99^{99}$ also ends with a $9.$ Adding them together yields a number that ends in $8.$

Thus, D is the correct answer.