Question

If $x^2$ is divisible by 216, what is the smallest possible value for positive integer x ?

• A. $72$
• B. $12$
• C. $36$
• D. $6$

Answer 1

Answer: (D)

The prime box of $x^2$ contains the prime factors of 216, which are 2, 2, 2, 3, 3, and 3. You know that the prime factors of $x^2$ should be the prime factors of x appearing in sets of two, or pairs. Therefore, you should distribute the prime factors of $x^2$ into two columns to represent the prime factors of x, as shown to the right.

There is a complete pair of two 2's in the prime box of $x^2$, so x must have a factor of 2. However, there is a third 2 left over. That additional factor of 2 must be from x as well, so assign it to one of the component x columns. Also, there is a complete pair of two 3's in the prime box of $x^2$, so x must have a factor of 3. However, there is a third 3 left over. That additional factor of 3 must be from x as well, so assign it to one of the component x columns. Thus, x has 2, 3, 2, and 3 in its prime box, so x must be a positive multiple of 36.