2026 AMC 10A Problem 11

The least common multiple (LCM) of a group of numbers is a multiple of every number in the group. So if one of the six numbers is the LCM, that number must be a multiple of all the others. Start by asking which number could play that role.

1. The LCM of the five known numbers. Factor them:
$$2 = 2, \quad 5 = 5, \quad 24 = 2^3 \cdot 3, \quad 60 = 2^2 \cdot 3 \cdot 5, \quad 240 = 2^4 \cdot 3 \cdot 5.$$
Take the highest power of each prime: $2^4 \cdot 3 \cdot 5 = 240$. So $240$ is already a multiple of $2$, $5$, $24$ and $60$.

1. Who can be the LCM of all six? The LCM of all six must be a multiple of $240$, because $240$ is one of the six. The numbers $2, 5, 24, 60$ are too small to be multiples of $240$. That leaves two possibilities: the LCM is $n$, or the LCM is $240$.

3. Case 1: the LCM is $n$. Then $n$ is a multiple of $240$. (Any multiple of $240$ is automatically a multiple of $2, 5, 24, 60$ too.) The three-digit multiples of $240$ are
$$240, \quad 480, \quad 720, \quad 960.$$
The six numbers must be different, so $n$ cannot be $240$. This case gives $n = 480, 720, 960$.

4. Case 2: the LCM is $240$. Then $240$ must be a multiple of $n$; in other words, $n$ divides $240$. List the divisors of $240$ in pairs that multiply to $240$:
$$1 \cdot 240, \quad 2 \cdot 120, \quad 3 \cdot 80, \quad 4 \cdot 60, \quad 5 \cdot 48, \quad 6 \cdot 40, \quad 8 \cdot 30, \quad 10 \cdot 24, \quad 12 \cdot 20, \quad 15 \cdot 16.$$
The only three-digit divisors are $120$ and $240$, and $240$ is not allowed again. This case gives $n = 120$.

1. Check Case 2. With $n = 120$, Gary's numbers are $2, 5, 24, 60, 120, 240$, and $240$ is a multiple of each one, so $240$ is their LCM. Good.

6. Add. The possible values are $120, 480, 720, 960$:
$$120 + 480 + 720 + 960 = 2280.$$

The answer is $\boxed{\textbf{(D)}\ 2280}$.

Saying "one of the numbers is the LCM of the group" is the same as saying "one number in the group is a multiple of all the others." Since the LCM can never be smaller than the largest number, only the biggest numbers are candidates, and here that leaves just two cases. Whenever a problem says the LCM (or the GCD) is one of the numbers, look at the largest (or smallest) numbers first. Also watch the word "distinct": it is what rules out $n = 240$.