HMMT November

I claim the answer is 120. First, note that $120=2^{3} \cdot 3 \cdot 5$, so the sum of divisors is $(1+2+4+8)(1+3)(1+5)=15 \cdot 4 \cdot 6=360$. Thus, we need to split the divisors into groups summing to 120 . But then we can just take $\{120\},\{20,40,60\},\{1,2,3,4,5,6,8,10,12,15,24,30\}$. Thus, 120 works. Now we need to show 120 is the lowest. Let $s(n)$ be the sum of divisors. Since $n$ will be in one of the piles, we need $s(n) \geq 3 n$. First, we claim that $n$ must have at least 3 distinct prime divisors. Surely, if it had 2 distinct prime divisors, say $p$ and $q$, so that $n=p^{a} q^{b}$, then the sum of divisors is $$\left(1+p+p^{2}+\ldots+p^{a}\right)\left(1+q+q^{2}+\ldots+q^{b}\right)=p^{a} q^{b}\left(1+\frac{1}{p}+\ldots+\frac{1}{p^{a}}\right)\left(1+\frac{1}{q}+\ldots+\frac{1}{q^{b}}\right)$$ However, the expression $1+\frac{1}{p}+\ldots+\frac{1}{p^{a}}$ is maximized when $p$ is minimized, and further, as $a$ is finite must be at most $\frac{1}{1-\frac{1}{p}}=\frac{p}{p-1}$. Thus, the sum of divisors is less than $$p^{a} q^{b} \frac{p}{p-1} \frac{q}{q-1} \leq n \cdot 2 \cdot \frac{3}{2}=3 n$$ Thus, $n$ can't have 2 distinct prime divisors and must have at least 3 distinct prime divisors. As we already discovered 120 works, we need not worry about 4 distinct prime divisors, as the value of $n$ would be at least $2 \cdot 3 \cdot 5 \cdot 7=210$. We now work through the numbers with 3 distinct divisors. If 2 is not one of them, then the only number that works is $105=3 \cdot 5 \cdot 7$, which has a sum of divisors that is not large enough. Therefore, 2 must be a prime divisor of $n$. Additionally, if 3 is not a divisor, then our options are $2 \cdot 5 \cdot 7$ and $2 \cdot 5 \cdot 11$, which also do not work. Therefore, 3 must also be a prime divisor. Then, if 5 is not a prime divisor, then if $n$ is $2 \cdot 3 \cdot p$, it has a sum of divisors of $(1+2)(1+3)(1+p)=n \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{p+1}{p}$, which is only at least $3 n$ if $p$ is exactly 2 , which is not feasible. Additionally, if we use $2^{2}$, then the sum of divisors is $(1+2+4)(1+3)(1+p)=n \cdot \frac{7}{4} \cdot \frac{4}{3} \cdot \frac{p}{p+1}$, so $\frac{p+1}{p}>\frac{9}{7} \Longrightarrow p<4.5$, which also can't happen. Further, we can't have $3^{2}$ be a divisor of $n$ as $2 \cdot 3^{2} \cdot 5$ is the only value less than 120 with this, and that also does not work. Lastly, we just need to check $2^{3} \cdot 3 \cdot p$, which has a sum of divisors of $(1+2+4+8)(1+3)(1+p)=n \cdot \frac{15}{8} \cdot \frac{4}{3} \cdot \frac{p+1}{p}=n \cdot \frac{5}{2} \cdot \frac{p}{p+1}$, so $p=5$ and that works. This means that $n=120$ is the smallest value for which $s(n) \geq 3 n$, and thus is our answer.