, 2015

First, we calculate the value of $72\left(\frac{3}{2}\right)^n$ for each integer from $n=-3$ to $n=4$, inclusive: $$\begin{aligned} 72\left(\tfrac{3}{2}\right)^{-3} &= 72 \cdot \tfrac{2^3}{3^3} = 72\cdot\tfrac{8}{27} = \tfrac{64}{3} \\ 72\left(\tfrac{3}{2}\right)^{-2} & = 72 \cdot \tfrac{2^2}{3^2} = 72\cdot\tfrac{4}{9} = 32\\ 72\left(\tfrac{3}{2}\right)^{-1} &= 72 \cdot \tfrac{2^1}{3^1} = 72\cdot\tfrac{2}{3} = 48 \\ 72\left(\tfrac{3}{2}\right)^{0} &= 72 \cdot 1 = 72 \\ 72\left(\tfrac{3}{2}\right)^{1} &= 72 \cdot \tfrac{3^1}{2^1} = 72\cdot\tfrac{3}{2} = 108 \\ 72\left(\tfrac{3}{2}\right)^{2} &= 72 \cdot \tfrac{3^2}{2^2} = 72\cdot\tfrac{9}{4} = 162 \\ 72\left(\tfrac{3}{2}\right)^{3} &= 72 \cdot \tfrac{3^3}{2^3} = 72\cdot\tfrac{27}{8} = 243\\ 72\left(\tfrac{3}{2}\right)^{4} &= 72 \cdot \tfrac{3^4}{2^4} = 72\cdot\tfrac{81}{16} = \tfrac{729}{2}\\ \end{aligned}$$ 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 $72\left(\tfrac{3}{2}\right)^{4} = \frac{729}{2}$, 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 $$72\left(\tfrac{3}{2}\right)^n = 3^2 \cdot 2^3 \cdot 3^n \cdot 2^{-n} = 3^2 3^n 2^3 2^{-n} = 3^{2+n} 2^{3-n}$$ 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.