2017 AMC 10B Problem 6

This is a "largest number" question, so it has two halves: show that some number of blocks fits, then show that one more cannot.

Part 1: five blocks cannot fit.

1. The box has volume $3 \cdot 2 \cdot 3 = 18$ cubic inches.
2. Each block has volume $2 \cdot 2 \cdot 1 = 4$ cubic inches.
3. Five blocks would need $5 \cdot 4 = 20$ cubic inches, more than the box has. So at most $4$ blocks fit.

Part 2: four blocks do fit.

1. Think of the box as $3$ inches wide, $2$ inches deep and $3$ inches tall.
2. Lay a block flat: $2$ wide, $2$ deep, $1$ tall. Stack three of them in the left part of the box. Together they fill a space $2$ wide, $2$ deep and $3$ tall.
3. What is left is a slab $1$ inch wide, $2$ deep and $3$ tall.
4. Turn the fourth block on its side so it is $1$ wide, $2$ deep and $2$ tall. It fits in the slab, with $1$ inch of height to spare.

Four blocks fit and five cannot, so the answer is $\boxed{\textbf{(B)}\ 4}$.

Packing questions always have two halves: a volume count that says "no more than this," and an actual arrangement that reaches that number. When the two agree, you are done. The blocks do not all have to face the same way; turning one of them is often what uses up the leftover space.

Upper bound: the box has volume $3 \cdot 2 \cdot 3 = 18$ cubic inches and each block has volume $2 \cdot 2 \cdot 1 = 4$. Since $4 \cdot 5 = 20 > 18$, at most $4$ blocks fit.

Construction: think of the box as $3$ wide, $2$ deep, $3$ tall. Lay a block flat so its $2 \times 2$ face covers a $2$ wide by $2$ deep footprint and it is $1$ tall; stack three of these to fill a $2 \times 2 \times 3$ column. The remaining space is a $1 \times 2 \times 3$ slab. Stand the fourth block on its edge: $1$ wide, $2$ deep, $2$ tall. It fits in the slab with one inch of height to spare.

Four blocks fit and five cannot, so the answer is $\boxed{\textbf{(B)}\ 4}$.