Competition problem — Number theory

Library / /142 of 144

Number theory Difficulty 3.0 AMC 10/12 Find the answer

For how many pairs $(m, n)$ with $m$ and $n$ integers satisfying $1 \leq m \leq 100$ and $101 \leq n \leq 205$ is $3^{m}+7^{n}$ divisible by 10?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The units digits of powers of 3 cycle $3,9,7,1$ and the units digits of powers of 7 cycle $7,9,3,1$ . For $3^{m}+7^{n}$ to be divisible by 10, one of the following must be true: units digit of $3^{m}$ is 3 and $7^{n}$ is 7, or 9 and 1, or 7 and 3, or 1 and 9. The number of possible pairs $(m, n)$ is $27 \times 25+26 \times 25+26 \times 25+26 \times 25=2625$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.