What is the sum of the tens digit and the ones digit of the integer form of $(2+3)^{23}$?
Simplify $(2+3)^{23}=5^{23}$. Since the ones digit of $5\times5$ is 5, the ones digit of $5^n$ is 5 for any positive integer $n$. Similarly, since the tens digit of $25\times5$ is 2 (and the ones digit is 5), the tens digit of $5^n$ is 2 for all positive integers $n\ge2$. Therefore, the sum of the tens digit and the ones digit of $(2+3)^{23}$ is $2+5=\boxed{7}$.
7
Level 2
Number Theory
Problem: What is the sum of the tens digit and the ones digit of the integer form of $(2+3)^{23}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of the tens digit and the ones digit of the integer form of $(2+3)^{23}$, first simplify the expression to $5^{23}$. Next, determine the last two digits of $5^{23}$. 

The powers of 5 exhibit a pattern in their last two digits. Specifically, for any integer $n \geq 2$, the last two digits of $5^n$ are always 25. This can be reasoned as follows: 

- $5^1 = 5$ (last two digits: 05)
- $5^2 = 25$ (last two digits: 25)
- $5^3 = 125$ (last two digits: 25)
- $5^4 = 625$ (last two digits: 25)
- Continuing this pattern, we see that for $n \geq 2$, $5^n$ consistently ends in 25.

Since $23 \geq 2$, we conclude that the last two digits of $5^{23}$ are also 25. 

Now, identify the tens digit and the ones digit from the last two digits. The tens digit is 2, and the ones digit is 5. 

Finally, calculate the sum of the tens digit and the ones digit: 

2 + 5 = 7. 

Thus, the final answer is 7.<|END_OF_LATENT|> The final answer is $7$.