### Instruction: What is the smallest base-10 integer that can be represented as $AA_5$ and $BB_7$, where $A$ and $B$ are valid digits in their respective bases?### Response: To find the smallest base-10 integer that can be represented as both $AA_5$ and $BB_7$, I need to find the smallest values of $A$ and $B$ that satisfy this condition.
I also need to convert $AA_5$ and $BB_7$ to base-10, so I can compare them.
To convert $AA_5$ to base-10, I can use the place value method: $AA_5 = A \times 5^1 + A \times 5^0 = 6A$.
Similarly, to convert $BB_7$ to base-10, I can use the same method: $BB_7 = B \times 7^1 + B \times 7^0 = 8B$.
Now I need to find the smallest values of $A$ and $B$ that make $6A = 8B$.
I can start by trying $A = 1$, which gives me $6 = 8B$.
This implies that $B = 0.75$, which is not a valid digit in base-7.
So I need to increase $A$ by one and try again.
If $A = 2$, then $6A = 12 = 8B$.
This implies that $B = 1.5$, which is also not a valid digit in base-7.
So I need to increase $A$ by one more and try again.
If $A = 3$, then $6A = 18 = 8B$.
This implies that $B = 2.25$, which is still not a valid digit in base-7.
So I need to increase $A$ by one more and try again.
If $A = 4$, then $6A = 24 = 8B$.
This implies that $B = 3$, which is a valid digit in base-7.
So I have found the smallest values of $A$ and $B$ that make $6A = 8B$: $A = 4$ and $B = 3$.
This means that the smallest base-10 integer that can be represented as both $AA_5$ and $BB_7$ is $24$.