2008 iTest Problems/Problem 25

Problem

A cube has edges of length $120$ cm. The cube gets chopped up into some number of smaller cubes, all of equal size, such that each edge of one of the smaller cubes has an integer length. One of those smaller cubes is then chopped up into some number of even smaller cubes, all of equal size. If the edge length of one of those even smaller cubes is $n$ cm, where $n$ is an integer, find the number of possible values of $n$ .

Solution

The prime factorization of $120$ is $2^3 \cdot 3 \cdot 5.$ Note that $n$ must be an integer, and since the cube is chopped up into smaller cubes of integral length, $n$ must be one of the $4 \cdot 2 \cdot 2 = 16$ factors of $120$ . However, since the cube is divided twice, we have to divide 120 by two numbers greater than $1,$ so the numbers $120, 60, 40,$ and $24$ would not work. Thus, there are $16 - 4 = \boxed{12}$ possible values of $n$ .

2008 iTest (Problems)
Preceded by: Problem 24	Followed by: Problem 26
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2008 iTest Problems/Problem 25

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