AMC 10 Step by Step

2008 AMC 10B Problem 4

P1-10~1 minPrint

A semipro baseball league has teams with 2121 players each. League rules state that a player must be paid at least $15,000\$15,000 and that the total of all players' salaries for each team cannot exceed $700,000.\$700,000. What is the maximum possible salary, in dollars, for a single player?

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Key insight

To make one salary as large as possible, pay everyone else the smallest amount allowed. The star gets whatever is left of the 700,000.

Solution

The team has a fixed amount of money. Every dollar paid to someone else is a dollar the star cannot get, so pay everyone else as little as the rules allow.

1. Besides the star there are 21−1=2021 - 1 = 20 players.
2. Each must get at least 15,00015{,}000. Paying each of them exactly that costs
20×15,000=300,000.20 \times 15{,}000 = 300{,}000.
3. The total cannot be more than 700,000700{,}000, so the star can get at most
700,000−300,000=400,000.700{,}000 - 300{,}000 = 400{,}000.
4. This really happens: twenty players at 15,00015{,}000 and one at 400,000400{,}000 add to exactly 700,000700{,}000, and everyone gets at least the minimum.

The answer is (C) 400,000\boxed{\textbf{(C)}\ 400,000}.

Why this works

When a total has a ceiling and each piece has a floor, one piece is biggest when all the other pieces sit at their floor. So "make one as large as possible" becomes "make all the others as small as possible", and the problem is one subtraction. Be careful to count the others, not everyone: 2020 players, not 2121.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The team's payroll is at most 700,000700{,}000. To leave as much as possible for one player, every one of the other 2020 players should earn the minimum, 15,00015{,}000:
20⋅15,000=300,000.20 \cdot 15{,}000 = 300{,}000.
The remaining money goes to the highest-paid player:
700,000−300,000=400,000.700{,}000 - 300{,}000 = 400{,}000.
This is attainable (twenty players at 15,00015{,}000 and one at 400,000400{,}000 totals exactly 700,000700{,}000), so it is the maximum.

The answer is (C) 400,000\boxed{\textbf{(C)}\ 400,000}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.