### Instruction: What is the units digit of $19^{19}+99^{99}$?### Response: I need to find the units digit of this sum, which means I only care about the units digits of the terms being added.
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.