2000 AMC 10 Problem 20

Attempt Problem 20 of the 2000 AMC 10 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 10 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

Let A,A, M,M, and CC be nonnegative integers such that A+M+C=10.A + M + C = 10. What is the maximum value of AMC+AM+MC+CA? \begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A? \end{aligned}

4949

5959

6969

7979

8989

Answer: C
Concepts:factoringoptimization
Difficulty rating: 1820
Solution:

Notice that AMC+AM+MC+CA=(A+1)(M+1)(C+1)(A+M+C)1=(A+1)(M+1)(C+1)11. \begin{gathered} A \cdot M \cdot C + AM + MC + CA \\ = (A+1)(M+1)(C+1) \\ {}- (A + M + C) - 1 \\ = (A+1)(M+1)(C+1) - 11. \end{gathered}

We maximize a product of three positive integers summing to 13.13. The most balanced split is 4,4,5,4, 4, 5, giving 445=80.4 \cdot 4 \cdot 5 = 80.

The maximum is 8011=69.80 - 11 = 69.

Thus, the correct answer is C.

← Problem 19#19
Full Exam

Problem 20 in Other Years