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2006 CEMC Pascal Problems/Problem 3

Problem

How many positive whole numbers, including $1$ and $18$, divide exactly into $18$?

$\text{ (A) }\ 3 \qquad\text{ (B) }\ 4 \qquad\text{ (C) }\ 5 \qquad\text{ (D) }\ 6 \qquad\text{ (E) }\ 7$

Solution 1

The numbers that divide exactly into $18$ are just the factors of $18$. This means that we can use the prime factorization of $18$ to find how many factors of $18$ there are.

$18 = 2 \times 3^{2}$, and we can have $2$ appear once or zero times in each factor ($2$ possibilities), and we can have $3$ appear $0$ to $2$ times ($3$ possibilities).

This means that the number of factors of $18$ is $2 \times 3 = \boxed {\textbf {(D) } 6}$

~anabel.disher

Solution 2

Since there aren't that many factors of $18$, we can simply list all of the factors:

$1, 2, 3, 6, 9, 18$

There are $\boxed {\textbf {(D) } 6}$ factors.

~anabel.disher

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