# 1996 AJHSME Problems/Problem 1

## Problem

How many positive factors of 36 are also multiples of 4? $\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

## Solution

The factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18,$ and $36$.

The multiples of $4$ up to $36$ are $4, 8, 12, 16, 20, 24, 28, 32$ and $36$.

Only $4, 12$ and $36$ appear on both lists, so the answer is $3$, which is option $\boxed{B}$.

## Solution 2 $36 = 4^1 \cdot 3^2$. All possible factors of $36$ will be here, except for ones divisible by $2$ and not by $4$. $(1+1)\cdot (2+1) = 6$. Subtract factors not divisible by $4$, which are $1$, $3^1$, and $3^2$. $6-3=3$, which is $\boxed{B}$.

## Solution 3

Divide $36$ by $4$, and the remaining factors, when multiplied by $4$, will be factors of $36$. $36 \div 4 = 9$, which has $3$ factors, giving us option $\boxed{B}$.

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