6. Knowing that the least common multiple of two natural numbers is equal to 240, what is the smallest possible value of their sum?
(A) 34
(B) 32
(C) 31
(D) 30
(E) 38
Problem 6. The correct answer is (C).

Let's call $a$ and $b$ two natural numbers that satisfy the requirement. We can assume that $a$ and $b$ are coprime. Indeed, if they were not, there would be at least one factor $p$ that divides both. By dividing by $p$ the one between $a$ and $b$ where such a factor $p$ appears with the least multiplicity (for example $b$), we would find that the least common multiple (LCM) between $a$ and $b / p$ would be the same as before, but the sum $a + b / p$ would be less than $a + b$ (contrary to the initial hypothesis). The numbers $a$ and $b$ are therefore coprime, and their LCM will thus be given by the product $a \cdot b$. It is therefore a matter of minimizing the sum $a + b$ among the pairs of natural numbers that are coprime and whose product is equal to 240. We can list the pairs of natural numbers that are coprime and whose product is $240 = 2^4 \cdot 3 \cdot 5$, simply by looking at how the three factors $\{2^4, 3, 5\}$ can be divided into two disjoint groups: $1 \cdot (2^4 \cdot 3 \cdot 5)$, $3 \cdot (5 \cdot 2^4)$, $5 \cdot (3 \cdot 2^4)$, $2^4 \cdot (3 \cdot 5)$. The respective sums are $1 + 240$, $3 + 80$, $5 + 48$, and $16 + 15$, among which the last one is the smallest.