2017/18 AHSMC Part 1 Problem 16

We have $n \mid 24^{240}$, then $n = 2^a 3^b$, with $a$ and $b$ integers. Thus $n$ has $(a + 1)(b + 1)$ divisors. Pairing them in pairs of the form $\bigl(d, \dfrac{n}{d}\bigr)$ we obtain that their product is $n^{\frac{(a+1)(b+1)}{2}}$. Therefore, $$24^{240} = n^{\frac{(a+1)(b+1)}{2}},$$ and hence $$2^{3 \cdot 240} 3^{240} = 2^{a \frac{(a+1)(b+1)}{2}} 3^{b \frac{(a+1)(b+1)}{2}}.$$ From this equation one obtains $$240 = b \frac{(a+1)(b+1)}{2}, \qquad 3 \cdot 240 = a \frac{(a+1)(b+1)}{2}$$ from which, taking the ratio, we get $a = 3b$ and hence, $$480 = b(b+1)(3b+1).$$ Since the RHS is increasing this equation has at most one positive integer solution. Moreover, we have $$480 = b(b+1)(3b+1) > 3b^3 \Rightarrow 160 > b^3$$ and hence $b \le 5$. By testing, $b = 5$ gives $$b(b+1)(3b+1) = 5 \cdot 6 \cdot 16 = 10 \cdot 6 \cdot 8 = 480.$$ Hence $b = 5$, $a = 15$ and $n = 2^{15} 3^5$. The answer is (B).