Difference between revisions of "2024 AMC 10A Problems/Problem 4"

Latest revision as of 19:19, 26 May 2025

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

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (Same as Solution 1 but Using 100=99+1)
5 Solution 4
6 Video Solution
7 Video Solution by Central Valley Math Circle
8 Video Solution by Math from my desk
9 Video Solution (⚡️ 55 sec solve ⚡️)
10 Video Solution by Pi Academy
11 Video Solution by Daily Dose of Math
12 Video Solution by FrankTutor
13 Video Solution 1 by Power Solve
14 Video Solution by SpreadTheMathLove
15 Video Solution by TheBeautyofMath
16 Video Solution by Dr. David
17 See also

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

Solution 4

The maximum $2$ -digit number is $99$ , but try $100$ . $2024 \div 100$ is a little more than $20$ , and the remainder is less than $100$ , by intuition, so there's $20 +$ the remainder $1 = \boxed{\textbf{(B) }21}$ .

~RandomMathGuy500

Video Solution

https://youtu.be/l3VrUsZkv8I ~MC

Video Solution by Central Valley Math Circle

https://youtu.be/aZqNhnTB_lQ

~mr_mathman

Video Solution by Math from my desk

https://www.youtube.com/watch?v=f6ogWpv56qw

Video Solution (⚡️ 55 sec solve ⚡️)

https://youtu.be/5nsOZQTcyMs

~Education, the Study of Everything

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 by FrankTutor

https://youtu.be/g2RxRsxrp2Y

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

Video Solution by TheBeautyofMath

For AMC 10: https://youtu.be/uKXSZyrIOeU?t=1084

For AMC 12: https://youtu.be/zaswZfIEibA?t=900

~IceMatrix

Video Solution by Dr. David

https://youtu.be/2onMJh_X2U4

2024 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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