Find the largest three-digit number that can be written in the form 3^m + 2^n, where m and n are positive integers.

Answer 1

3^m goes over 3 digits at 3^7
so m<7
let m = 6 to get 3^6 = 729
so that leaves 270 for 2^n
I know 2^8 = 256 , using 2^9 would put me over.

3^6 + 2^8 = 729+256 = 985

Answer 2

Since $3^7 > 1000$, we know that we only have to test 1 through 6 for $m$. For each value of $m$ from 1 through 5, the greatest possible $n$ is $n=9$, and the largest three-digit number we get from these is $3^5 + 2^9 = 243 + 512=755$. But when $m=6$, the largest three-digit number we get is $3^6 + 2^8 = 729 + 256 = \boxed{985}$.