Problem 583

We determine when $3^m + 7^n$ is divisible by 10 by looking at the units (ones) digits of $3^m + 7^n$.

To do this, we first look individually at the units digits of $3^m$ and $7^n$.

The units digits of powers of 3 cycle $3,9,7,1,3,9,7,1,\ldots$.

To see this, we note that the first few powers of 3 are $$3^1 = 3 \quad 3^2 = 9 \quad 3^3 = 27 \quad 3^4 = 81 \quad 3^5 = 243 \quad 3^6 = 729$$ Since the units digit of a product of integers depends only on the units digits of the integers being multiplied and we multiply by 3 to get from one power to the next, then once a units digit recurs in the sequence of units digits, the following units digits will follow the same pattern.

This means that the units digits of powers of 3 cycle every four powers of 3.

Therefore, of the 100 powers of 3 of the form $3^m$ with $1 \leq m \leq 100$, exactly 25 will have a units digit of 3, exactly 25 will have a units digit of 9, exactly 25 will have a units digit of 7, and exactly 25 will have a units digit of 1.

The units digits of powers of 7 cycle $7,9,3,1,7,9,3,1,\ldots$.

To see this, we note that the first few powers of 7 are $$7^1 = 7 \quad 7^2 = 49 \quad 7^3 = 343 \quad 7^4 = 2401 \quad 7^5 = 16\,807 \quad 7^6 = 117\,649$$ Using the same argument as above, the units digits of powers of 7 cycle every four powers of 7.

Since $101$ is 1 more than a multiple of 4, then the power $7^{101}$ is at the beginning of one of these cycles, and so the units digit of $7^{101}$ is a 7.

Therefore, of the 105 powers of 7 of the form $7^n$ with $101 \leq n \leq 205$, exactly 27 will have a units digit of 7, exactly 26 will have a units digit of 9, exactly 26 will have a units digit of 3, and exactly 26 will have a units digit of 1. (Here, 105 powers include 26 complete cycles of 4 plus one additional term.)

For $3^m+7^n$ to have a units digit of 0 (and thus be divisible by 10), one of the following must be true:

the units digit of $3^m$ is 3 (25 possible values of $m$) and the units digit of $7^n$ is 7 (27 possible values of $n$), or
the units digit of $3^m$ is 9 (25 possible values of $m$) and the units digit of $7^n$ is 1 (26 possible values of $n$), or
the units digit of $3^m$ is 7 (25 possible values of $m$) and the units digit of $7^n$ is 3 (26 possible values of $n$), or
the units digit of $3^m$ is 1 (25 possible values of $m$) and the units digit of $7^n$ is 9 (26 possible values of $n$).

The number of possible pairs $(m,n)$ is therefore $$27 \times 25 + 26 \times 25 + 26 \times 25 + 26 \times 25 = 25 \times (27+26+26+25) = 25 \times 105 = 2625$$