2000 AMC 10 Problem 20

1. Try some numbers. Call the expression $E = AMC + AM + MC + CA$.
- $A, M, C = 10, 0, 0$: $E = 0$.
- $5, 5, 0$: $E = 0 + 25 + 0 + 0 = 25$.
- $3, 3, 4$: $E = 36 + 9 + 12 + 12 = 69$.

Numbers close together seem to do best. Now let us prove $69$ is the most.
2. Spot a product. Multiply out $(A+1)(M+1)(C+1)$:
$$(A+1)(M+1)(C+1) = AMC + AM + MC + CA + A + M + C + 1.$$
The first four terms are $E$. The rest is $A + M + C + 1 = 10 + 1 = 11$. So
$$E = (A+1)(M+1)(C+1) - 11.$$
3. New question. Now we want three whole numbers $A+1$, $M+1$, $C+1$ (each at least $1$) that add to $13$ and have the biggest product.
4. Why equal is best. Suppose two of them differ by $2$ or more, like $6$ and $3$. Moving $1$ from the bigger to the smaller gives $5$ and $4$. The sum stays the same, but the product goes from $18$ to $20$. This always happens: moving $1$ from the bigger to the smaller raises the product. So at the maximum, no two of the numbers differ by $2$ or more.
5. Three whole numbers adding to $13$, all within $1$ of each other, must be $4, 4, 5$. Their product is $80$.
6. So the largest $E$ is $80 - 11 = 69$, reached at $A, M, C = 3, 3, 4$, as found in step 1.

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

An expression that looks like "most of" a multiplied-out product usually becomes a product once you add the missing terms. Here the missing terms were exactly $A + M + C + 1$, which is known. Then a fixed sum with the biggest product means the numbers should be as equal as possible, like a square having the most area of all rectangles with a given perimeter.

Trust the "as equal as possible" idea and compare nearby choices directly. $3, 3, 4$ gives $69$; $2, 4, 4$ gives $32 + 8 + 16 + 8 = 64$; $2, 3, 5$ gives $30 + 6 + 15 + 10 = 61$. The best is $69$, but the product argument above is what shows no other choice beats it.

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 $\boxed{\textbf{(C)}\ 69}$.