2024 AMC 10A Problems/Problem 4

The following problem is from both the 2024 AMC 10A #4 and 2024 AMC 12A #3, so both problems redirect to this page.

Problem

The number $2024$ 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?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$

Solution 1

Since we want the least number of two-digit numbers, we maximize the two-digit numbers by choosing as many $99$s as possible. Since $2024=99\cdot20+44\cdot1,$ we choose twenty $99$s and one $44,$ for a total of $\boxed{\textbf{(B) }21}$ two-digit numbers.

~MRENTHUSIASM

Solution 2

We claim the answer is $21$. This can be achieved by adding twenty $99$'s and a $44$. To prove that the answer cannot be less than or equal to $20$, we note that the maximum value of the sum of $20$ or less two digit numbers is $20 \cdot 99 = 1980$, which is smaller than $2024$, so we are done. Thus, the answer is $\boxed{\textbf{(B) }21}$.

~andliu766

Solution 3 (Same as solution 1 but Using 100=99+1)

$2024=100\cdot20+24$. Since $100=99+1$, $2024=(99+1)\cdot20+24=99\cdot20+1\cdot20+24=99\cdot20+44$. Therefore a total of $\boxed{\textbf{(B) }21}$ two-digit numbers are needed.

~woh123

Video Solution by Pi Academy

https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW

Video Solution by Daily Dose of Math

https://youtu.be/sEk9jQnMzfk

~Thesmartgreekmathdude

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=rWQoAYu7QsZP8ty4&t=407

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2024 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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