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 $\boxed{B}$ .

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by 1995 AJHSME Last Question	Followed by Problem 2
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1996 AJHSME Problems/Problem 1

Contents

Problem

Solution

Solution 2

Solution 3

See also