2017 AMC 10B Problem 6

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 $\boxed{\textbf{(B)}\ 4}$.

Packing questions have two halves: a volume (or other counting) bound that says "no more than this," and an explicit arrangement that reaches it. When the bound and the construction agree, you are done. Blocks need not all share the same orientation; rotating one is often the key to using leftover space.

• Stopping at 3 blocks after stacking three of them flat, without noticing the leftover 1 by 2 by 3 slab holds a fourth block turned on its side.
• Answering $\lfloor 18/4 \rfloor = 4$ by volume alone without confirming that four blocks can actually be arranged; here it happens to work, but the check is part of the argument.