2026 AMC 10A Problem 20

Side lengths that multiply to a number are built out of its prime factors, so factor first.

1. Factor the volume. Divide by small primes one at a time:
$$510510 = 2 \cdot 255255 = 2 \cdot 3 \cdot 85085 = 2 \cdot 3 \cdot 5 \cdot 17017 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 2431 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 221,$$
and $221 = 13 \cdot 17$. So
$$510510 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17,$$
seven different primes, each used exactly once.

1. Check with the two given boxes. $77 = 7 \cdot 11$, $78 = 2 \cdot 3 \cdot 13$ and $85 = 5 \cdot 17$: each prime lands in exactly one side. Likewise $2$, $3$ and $85085 = 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$.

3. A box is a way to split the primes into three groups. Each side is the product of the primes in its group, and each prime goes into exactly one side.
- "Side at least $2$" means no group is empty (an empty group would give a side of $1$).
- Different splits give different boxes: because each prime appears only once, you can read off which primes a side contains just by factoring it.

So we must count the ways to split $7$ different primes into $3$ non-empty groups, where the groups have no order.

4. First pretend the sides have names. Call them length, width and height. Each of the $7$ primes chooses one of the $3$ names, so there are
$$3^7 = 2187$$
ways. Some of these leave a side with no primes; remove those next.

5. Remove the ones with an empty side (inclusion–exclusion).
- Ways where the length gets no prime: each prime chooses between the other $2$ names, $2^7 = 128$ ways. The same is true for width and for height, so subtract $3 \cdot 128 = 384$.
- A way with two empty sides (all seven primes on one side) was subtracted twice, once for each of its empty sides. There are $3$ such ways (all on the length, all on the width, all on the height), so add $3$ back.
- No way has all three sides empty.

With named sides, the count is
$$2187 - 384 + 3 = 1806.$$

6. Now remove the names. A box does not care which side is called length. Its three sides are always three different numbers (different groups of primes have different products), so each box was counted once for each order of its sides, $3! = 6$ times:
$$\frac{1806}{6} = 301.$$

1. Test the method on a small case. With four primes, $2 \cdot 3 \cdot 5 \cdot 7 = 210$, one side must be a product of two primes and the others single primes, so there are $\binom{4}{2} = 6$ boxes. The method gives $\frac{3^4 - 3 \cdot 2^4 + 3}{6} = \frac{81 - 48 + 3}{6} = 6$. It agrees.

The answer is $\boxed{\textbf{(C)}\ 301}$.

When a number is a product of different primes, each factor is decided completely by which primes it takes, so a factoring question turns into "put different objects into groups". The standard plan is: count with named groups, use inclusion–exclusion to rule out empty groups, then divide by the number of orders. Dividing by $6$ is safe here only because no two sides can be equal; if a prime were repeated, two sides could match and those boxes would need separate care. (Aside: the number of ways to split $7$ different objects into $3$ non-empty unnamed groups is called a Stirling number of the second kind, and it equals $301$.)

Count by the sizes of the three groups. The sizes add to $7$ and each is at least $1$, so they are $5{+}1{+}1$, $4{+}2{+}1$, $3{+}3{+}1$ or $3{+}2{+}2$.

The total is $21 + 105 + 70 + 105 = 301$. When two groups have the same size, divide by $2$, because swapping those two groups gives the same box.