Japan Mathematical Olympiad

If a positive integer $l$ is a factor of a positive integer $n$, then $n/l$ is also a factor of $n$ and vice versa. Hence, if $\{a_1,a_2,\dots ,a_r\}$ is a listing of all the positive factors of $n$, then so is $\{n/a_1,n/a_2,\dots ,n/a_r\}$. So, if we denote by $N$ the product of all the positive factors of $n$, we obtain $$N^2=(a_1\cdot \frac{n}{a_1})\cdots (a_r\cdot \frac{n}{a_r})=n^r,$$ from which we conclude that $N=n^{r/2}$ holds.

Now, suppose that the product $N$ of all the positive factors of $n$ equals $24^{240}$, then since the prime factors of $24^{240}$ are $2$ and $3$, we see that the prime factors of $n$ must also be $2$ and $3$. Hence, we can write $n=2^a{3}^b$ for some pair of non-negative integers $a$ and $b$. Then, we can conclude that all of the positive factors of $n$ can be listed as $\{2^{a^{\prime }}{3}^{b^{\prime }}:0\le a^{\prime }\le a,0\le b^{\prime }\le b\}$ and there are exactly $(a+1)(b+1)$ numbers in this list. Hence, from the observation made above, we get that the product $N$ of all the factors of $n$ equals $$(2^a{3}^b{)}^{\frac{(a+1)(b+1)}{2}}=2^{\frac{a(a+1)(b+1)}{2}}{3}^{\frac{b(b+1)(a+1)}{2}},$$ and this number should equal $24^{240}=2^{720}{3}^{240}$. So, it remains for us to solve the simultaneous equations

$$\{\begin{array}{l}\frac{a(a+1)(b+1)}{2}=720\\ \frac{b(b+1)(a+1)}{2}=240\end{array}$$ Dividing both sides of the first equation by the corresponding sides of the second equation, we get $a=3b$, and substituting this into the second equation, we obtain $(3b+1)b(b+1)=480$, from which it follows that $$(3b+1)b(b+1)-480=(b-5)(3b^2+19b+96)=0$$ As the equation $3b^2+19b+96=0$ does not have a real root, we conclude that $b=5$ must hold, and therefore, $a=15$. Thus, $n=2^{15}{3}^5$ is the desired solution for this problem.