Let $S \subset \mathbb{N}$ be a set of positive integers whose product is 2021 times its sum.

a) Given that $S$ has five (or more) elements, what its minimum possible sum?

b) Given that $S$ has exactly four elements, what its minimum possible sum?

c) Given that $S$ has exactly three elements, what its minimum possible sum?
2021 is prime factorized as $43 \cdot 47$. There must therefore be at least one multiple of 43 and at least one multiple of 47 .

a) The answer is $105=47+43+7+5+3 \cdot 2 \cdot 43+47$ is already too much, so any smaller answer would have to involve both 43 and 47 . If $T$ is $S \backslash\{43,47\}$, the product of the elements of $T$ is thus $43+47=90$ more than the sum of elements of $T$. However, one can check that this cannot hold for any product under 105 .

b) The answer is $112=47+43+14+8$. We define $T$ similarly and let its elements be $a$ and $b$. We have $a+b+90=a b$, which can be rewritten as $(a-1)(b-1)=91=7 \cdot 13$. The minimum value of $a+b-2=(a-1)+(b-1)$ is thus $7+13=20$, giving an answer of $22+90$, as desired.

c) The answer is $240=94+86+60$. Let the numbers be $47 a, 43 b, x$, so that $47 a+43 b+x=a b x$. Solving for $x$ gives $x=47 a+43 b /(a b-1)$. In order to get less than 240 , we must have $a+b<240 / 43<6$. Checking all combinations gives the answer.