# 1992 AHSME Problems/Problem 4

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

If $a,b$ and $c$ are positive integers and $a$ and $b$ are odd, then $3^a+(b-1)^2c$ is $\text{(A) odd for all choices of c} \quad \text{(B) even for all choices of c} \quad\\ \text{(C) odd if c is even; even if c is odd} \quad\\ \text{(D) odd if c is odd; even if c is even} \quad\\ \text{(E) odd if c is not a multiple of 3; even if c is a multiple of 3}$

## Solution

Since 3 has no factors of 2, $3^a$ will be odd for all values of $a$. Since $b$ is odd as well, $b-1$ must be even, so $(b-1)^2$ must be even. This means that for all choices of $c$, $(b-1)^2c$ must be even because any integer times an even number is still even. Since an odd number plus an even number is odd, $3^a+(b-1)^2c$ must be odd for all choices of $c$, which corresponds to answer choice $\fbox{A}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 