Find the number of ordered pairs $(m,n)$ from the set of natural numbers from 1 to 20 such that $3^m + 7^n$ which is divisible by 10.

The answer is 100.

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For $3^{n}, n=1,2,3,4,...$ we have cycles of $3,9,7,1 \mod 10$

Similarly for $7^{m}, m=1,2,3,4,...$ we have cycles of $7,9,3,1 \mod 10$

For any value of $n$ , therefore, there is a corresponding value of $m$ that would make $3^{n}+7^{m}\equiv0 \mod10$

Choose any value $n$ , one $m$ in each cycle of the powers of $7$ will make the sum divisible by $10$ , as there are $4$ items in each cycle, $1$ in every $4$ values of $m$ is valid. There are $20$ possible values for $m$ so $5$ will work for each of the $20$ values of $n$ , so the answer is $5*20=\boxed{100}$