2000 AMC 10 Problem 20

The expression looks like most of an expanded product. Indeed,
$$(A+1)(M+1)(C+1) = AMC + AM + MC + CA + (A + M + C) + 1,$$
and since $A + M + C = 10$, the target expression equals
$$(A+1)(M+1)(C+1) - 11.$$

So we maximize a product of three positive integers $a = A+1$, $b = M+1$, $c = C+1$ with $a + b + c = 13$. If two of them differ by $2$ or more, say $a \ge c + 2$, replacing $(a, c)$ by $(a - 1, c + 1)$ keeps the sum and raises the product, because $(a-1)(c+1) = ac + (a - c) - 1 > ac$. So at the maximum all three are within $1$ of each other: $\{4, 4, 5\}$, product $80$.

Therefore the maximum is $80 - 11 = 69$, achieved at $(A, M, C) = (3, 3, 4)$: check $36 + 9 + 12 + 12 = 69$.

The answer is $\boxed{\textbf{(C)}\ 69}$.

A symmetric expression that is "almost" a product usually is one after adding the missing terms, and here the missing terms are exactly the given constraint plus $1$. Once it is a product with a fixed sum of factors, the smoothing argument (or AM-GM) says balanced factors win. This completing-the-product move is the same idea as Simon's Favoring Factoring Trick one dimension up.

Trust the balance heuristic and compute directly: $(3, 3, 4)$ gives $36 + 9 + 12 + 12 = 69$, while nearby triples such as $(2, 4, 4)$ give $32 + 8 + 16 + 8 = 64$ and $(2, 3, 5)$ gives $30 + 6 + 15 + 10 = 61$. The identity is what turns this hunch into a proof.

• Maximizing only the AMC term, or testing a couple of lopsided triples like (10,0,0) and (5,5,0) and stopping.
• Subtracting $10$ instead of $11$ after computing the product $80$, giving $70$, which is not a choice and should prompt a recheck.