AMC 10 Step by Step

2024 AMC 10A Problem 4

P1-10~1 minPrint

The number 20242024 is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

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Key insight

The biggest two-digit number is 99. Twenty of them make only 1980, which is short of 2024, but twenty 99s plus a 44 make exactly 2024.

Solution

To use as few numbers as possible, each number should be as large as possible. The largest two-digit number is 9999.

This is a "least possible" question, so it has two halves.

Part 1: twenty numbers are not enough.

  1. Twenty two-digit numbers add to at most 20⋅99=198020 \cdot 99 = 1980.
  2. But 19801980 is less than 20242024. So twenty numbers cannot reach 20242024, and fewer than twenty cannot either.

Part 2: twenty-one numbers are enough.

  1. After twenty 9999s we still need 2024−1980=442024 - 1980 = 44.
  2. 4444 is a two-digit number. So twenty 9999s and one 4444 add to exactly 20242024. That is 2121 numbers.

Twenty is too few and twenty-one works, so the least number is 2121.

The answer is (B) 21\boxed{\textbf{(B)}\ 21}.

Why this works

"Least number of pieces" questions always have two parts: show that one fewer piece can never be enough (use the biggest allowed piece to see how far they can reach), and then actually build the total with the number you claim. Filling up with the biggest pieces first is usually the way to build it.

Concise solution

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

Each two-digit number is at most 9999, so 2020 of them add to at most 20⋅99=198020\cdot 99 = 1980, which is less than 20242024. Twenty (or fewer) numbers cannot work.

Twenty-one can: 2024=20⋅99+442024 = 20\cdot 99 + 44, and 4444 is a two-digit number. So twenty 9999s and one 4444 do the job.

The answer is (B) 21\boxed{\textbf{(B)}\ 21}.

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. Also appeared as 2024 12A #3.