2024 AMC 10A Problem 4

To use as few numbers as possible, each number should be as large as possible. The largest two-digit number is $99$.

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 $20 \cdot 99 = 1980$.
2. But $1980$ is less than $2024$. So twenty numbers cannot reach $2024$, and fewer than twenty cannot either.

Part 2: twenty-one numbers are enough.

1. After twenty $99$s we still need $2024 - 1980 = 44$.
2. $44$ is a two-digit number. So twenty $99$s and one $44$ add to exactly $2024$. That is $21$ numbers.

Twenty is too few and twenty-one works, so the least number is $21$.

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

"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.

Each two-digit number is at most $99$, so $20$ of them add to at most $20\cdot 99 = 1980$, which is less than $2024$. Twenty (or fewer) numbers cannot work.

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

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