1. Let the two-digit number be $n$, so $10 \le n \le 99$.

2. We need: $3^n + 7^n \text{ is divisible by } 10$ That means: $3^n + 7^n \equiv 0 \pmod{10}$

3. Now look at powers of $3$ and $7$ modulo $10$.

For $3^n \pmod{10}$: $3,9,7,1,3,9,7,1,\dots$ This repeats with period $4$.

For $7^n \pmod{10}$: $7,9,3,1,7,9,3,1,\dots$ This also repeats with period $4$.

4. Add them according to $n \pmod{4}$:

• If $n \equiv 1 \pmod{4}$, then $3^n + 7^n \equiv 3+7=10 \equiv 0 \pmod{10}$
• If $n \equiv 2 \pmod{4}$, then $3^n + 7^n \equiv 9+9=18 \equiv 8 \pmod{10}$
• If $n \equiv 3 \pmod{4}$, then $3^n + 7^n \equiv 7+3=10 \equiv 0 \pmod{10}$
• If $n \equiv 0 \pmod{4}$, then $3^n + 7^n \equiv 1+1=2 \pmod{10}$

So the expression is divisible by $10$ exactly when $n$ is odd.

5. Therefore, we just count the odd two-digit numbers from $10$ to $99$. These are: $11,13,15,\dots,99$ This is an arithmetic progression with first term $11$, last term $99$, common difference $2$.

Number of terms: $\frac{99-11}{2}+1=\frac{88}{2}+1=44+1=45$

6. Hence, the total number of such two-digit numbers is: $\boxed{45}$