2024 AMC 10A Problems/Problem 4: Difference between revisions

Latest revision as of 20:07, 9 October 2025

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

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

Video Solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=684s

@@ Line 17: / Line 17: @@
 ~andliu766
-== Solution 3 (Same as solution 1 but Using 100=99+1)==
+== Solution 3 (Same as Solution 1 but Using 100=99+1)==
 <math>2024=100\cdot20+24</math>. Since <math>100=99+1</math>, <math>2024=(99+1)\cdot20+24=99\cdot20+1\cdot20+24=99\cdot20+44</math>. Therefore a total of <math>\boxed{\textbf{(B) }21}</math> two-digit numbers are needed.
 ~woh123
+== Solution 4 ==
+The maximum <math>2</math>-digit number is <math>99</math>, but try <math>100</math>. <math>2024 \div 100</math> is a little more than <math>20</math>, and the remainder is less than <math>100</math>, by intuition, so there's <math>20 +</math> the remainder <math>1 = \boxed{\textbf{(B) }21}</math>.
+~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
+<i>~Education, the Study of Everything </i>
 == Video Solution by Pi Academy ==
@@ Line 30: / Line 55: @@
 ~Thesmartgreekmathdude
+== Video Solution by FrankTutor ==
+https://youtu.be/g2RxRsxrp2Y
 == Video Solution 1 by Power Solve ==
@@ Line 37: / Line 66: @@
 https://www.youtube.com/watch?v=6SQ74nt3ynw
-==See also==
+==Video Solution by TheBeautyofMath==
-{{AMC10 box|year=2024|ab=A|num-b=3|num-a=5}}
+For AMC 10: https://youtu.be/uKXSZyrIOeU?t=1084
-{{AMC12 box|year=2024|ab=A|num-b=2|num-a=4}}
+For AMC 12: https://youtu.be/zaswZfIEibA?t=900
+~IceMatrix
+==Video Solution by Dr. David==
+https://youtu.be/2onMJh_X2U4
+==Video Solution by TheNeuralMathAcademy==
+https://www.youtube.com/watch?v=4b_YLnyegtw&t=684s
+==See Also==
+{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
+* [[AMC 10]]
+* [[AMC 10 Problems and Solutions]]
+* [[Mathematics competitions]]
+* [[Mathematics competition resources]]
 {{MAA Notice}}

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