For how many integers $n$ is $72\left(\frac{3}{2}\right)^{n}$ equal to an integer?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

## Part C: Each correct answer is worth 8.
First, we calculate the value of $72\left(\frac{3}{2}\right)^{n}$ for each integer from $n=-3$ to $n=4$, inclusive:

$$
\begin{array}{lc}
72\left(\frac{3}{2}\right)^{-3}=72 \cdot \frac{2^{3}}{3^{3}}=72 \cdot \frac{8}{27}=\frac{64}{3} & 72\left(\frac{3}{2}\right)^{-2}=72 \cdot \frac{2^{2}}{3^{2}}=72 \cdot \frac{4}{9}=32 \\
72\left(\frac{3}{2}\right)^{-1}=72 \cdot \frac{2^{1}}{3^{1}}=72 \cdot \frac{2}{3}=48 & 72\left(\frac{3}{2}\right)^{0}=72 \cdot 1=72 \\
72\left(\frac{3}{2}\right)^{1}=72 \cdot \frac{3^{1}}{2^{1}}=72 \cdot \frac{3}{2}=108 & 72\left(\frac{3}{2}\right)^{2}=72 \cdot \frac{3^{2}}{2^{2}}=72 \cdot \frac{9}{4}=162 \\
72\left(\frac{3}{2}\right)^{3}=72 \cdot \frac{3^{3}}{2^{3}}=72 \cdot \frac{27}{8}=243 & 72\left(\frac{3}{2}\right)^{4}=72 \cdot \frac{3^{4}}{2^{4}}=72 \cdot \frac{81}{16}=\frac{729}{2}
\end{array}
$$

Therefore, there are at least 6 integer values of $n$ for which $72\left(\frac{3}{2}\right)^{n}$ is an integer, namely $n=-2,-1,0,1,2,3$.

Since 6 is the largest possible choice given, then it must be the correct answer (that is, it must be the case that there are no more values of $n$ that work).

We can justify this statement informally by noting that if we start with $72\left(\frac{3}{2}\right)^{4}=\frac{729}{2}$, then making $n$ larger has the effect of continuing to multiply by $\frac{3}{2}$ which keeps the numerator odd and the denominator even, and so $72\left(\frac{3}{2}\right)^{n}$ is never an integer when $n>3$. A similar argument holds when $n<-2$.

We could justify the statement more formally by re-writing

$$
72\left(\frac{3}{2}\right)^{n}=3^{2} \cdot 2^{3} \cdot 3^{n} \cdot 2^{-n}=3^{2} 3^{n} 2^{3} 2^{-n}=3^{2+n} 2^{3-n}
$$

For this product to be an integer, it must be the case that each of $3^{2+n}$ and $2^{3-n}$ is an integer. (Each of $3^{2+n}$ and $2^{3-n}$ is either an integer or a fraction with numerator 1 and denominator equal to a power of 2 or 3 . If each is such a fraction, then their product is less than 1 and so is not an integer. If exactly one is an integer, then their product equals a power of 2 divided by a power of 3 or vice versa. Such a fraction cannot be an integer since powers of 3 cannot be "divided out" of powers of 2 and vice versa.)

This means that $2+n \geq 0$ (and so $n \geq-2$ ) and $3-n \geq 0$ (and so $n \leq 3$ ).

Therefore, $-2 \leq n \leq 3$. The integers in this range are the six integers listed above.

ANSWER: (E)