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

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by 1995 AJHSME Last Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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

Page

Toolbox

Search

1996 AJHSME Problems/Problem 1

Contents

Problem

Solution

Solution 2

See also