What is the units digit of $19^{19}+99^{99}$?
I can ignore the tens, hundreds, and other digits of the terms, since they will not affect the units digit of the sum. To find the units digit of a power, I can look for a pattern in the units digits of successive powers of the same base. For example, for $19^{19}$, I can start with $19^1$, which has a units digit of 9. Then, $19^2$ has a units digit of 1, since 9 times 9 is 81. Then, $19^3$ has a units digit of 9 again, since 9 times 1 is 9. I notice that the units digit of $19^n$ alternates between 9 and 1, depending on whether n is odd or even. Since $19^{19}$ has an odd exponent, its units digit is 9, the same as $19^1$. Similarly, for $99^{99}$, I can start with $99^1$, which has a units digit of 9. Then, $99^2$ has a units digit of 1, since 9 times 9 is 81. Then, $99^3$ has a units digit of 9 again, since 9 times 1 is 9. I notice that the units digit of $99^n$ also alternates between 9 and 1, depending on whether n is odd or even. Since $99^{99}$ has an odd exponent, its units digit is 9, the same as $99^1$. Therefore, the units digit of $19^{19}+99^{99}$ is the same as the units digit of $9+9$, which is 8. 