Product Bricks

Let $x$ be the number in the brick left of brick $A$ , and $y$ be the number in the brick right of brick $C$ . Then the brick above bricks $x$ and $A$ contains $x A$ , the brick above bricks $A$ and $B$ contains $A B$ , the brick above these last two bricks contains $x A^2 B$ , etc. Continuing to the top produces the completed pyramid below:

We need to find the lowest value of $k$ such that $k! = x A^4 B^6 C^4 y$ for some $x, A, B, C, y \in \mathbb{N}_{>1}$ . Writing the prime factorizations of the first few factorials,

• $2! = 2$

• $3! = 3$

$\qquad \; \; \; \vdots$

• $8! = 2^7 \cdot 3^2 \cdot 5 \cdot 7$

• $9! = 2^7 \cdot 3^4 \cdot 5 \cdot 7$

• $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$

• $11! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 \cdot 11$

• $12! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$

We see that $12!$ 's expansion is the first one that has exponents high enough to make two $4$ th powers and a $6$ th. To accomplish this, we split $2^{10}$ into $2^6 \cdot 2^4$ and $3^5$ into $3^4 \cdot 3$ . Then we let $A = B = 2$ and $C = 3$ ; the remaining factors of $3, 5, 5, 7$ and $11$ can be distributed among $x$ and $y$ in any fashion, e.g. $x = 3 \cdot 5 \cdot 7 = 105, y = 5 \cdot 11 = 55$ . Then $k! = 12! = 105 \cdot 2^4 \cdot 2^6 \cdot 3^4 \cdot 55$ .

The only variation allowed by the above solution (other than $x$ and $y$ ) is which of $A, B$ or $C$ equals $3$ , and regardless of that we will have $A \cdot B \cdot C = 2^2 \cdot 3 = \boxed{12}$