The total number of two digit numbers $$'n'$$, such that $$3^n + 7^n$$ is a multiple of 10, is ______.

We need to find the number of two-digit numbers $$n$$ such that $$3^n + 7^n$$ is a multiple of 10. For this, we need $$3^n + 7^n \equiv 0 \pmod{10}$$.

We examine the last digits of powers of 3 and 7. The cycle of last digits of $$3^n$$ is: $$3, 9, 7, 1$$ (period 4). The cycle of last digits of $$7^n$$ is: $$7, 9, 3, 1$$ (period 4).

Adding the last digits for each $$n \bmod 4$$: when $$n \equiv 1 \pmod{4}$$, last digits sum to $$3 + 7 = 10$$; when $$n \equiv 2 \pmod{4}$$, last digits sum to $$9 + 9 = 18$$; when $$n \equiv 3 \pmod{4}$$, last digits sum to $$7 + 3 = 10$$; when $$n \equiv 0 \pmod{4}$$, last digits sum to $$1 + 1 = 2$$.

So $$3^n + 7^n$$ is divisible by 10 when the last digit sum is 10 or 20, etc. From above, this happens when $$n \equiv 1$$ or $$3 \pmod{4}$$, i.e., when $$n$$ is odd.

The two-digit numbers range from 10 to 99. The odd two-digit numbers are 11, 13, 15, ..., 99. Their count is $$\frac{99 - 11}{2} + 1 = 45$$.

Therefore, the total number of such two-digit numbers is $$45$$.