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

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{{duplicate|[[2024 AMC 10A Problems/Problem 4|2024 AMC 10A #4]] and [[2024 AMC 12A Problems/Problem 3|2024 AMC 12A #3]]}}
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== Problem ==
  
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The number <math>2024</math> 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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<math>\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24</math>
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== Solution 1 ==
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Since we want the least number of two-digit numbers, we maximize the two-digit numbers by choosing as many <math>99</math>s as possible. Since <math>2024=99\cdot20+44\cdot1,</math> we choose twenty <math>99</math>s and one <math>44,</math> for a total of <math>\boxed{\textbf{(B) }21}</math> two-digit numbers.
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~MRENTHUSIASM
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== Solution 2 ==
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We claim the answer is <math>21</math>. This can be achieved by adding twenty <math>99</math>'s and a <math>44</math>. To prove that the answer cannot be less than or equal to <math>20</math>, we note that the maximum value of the sum of <math>20</math> or less two digit numbers is <math>20 \cdot 99 = 1980</math>, which is smaller than <math>2024</math>, so we are done. Thus, the answer is <math>\boxed{\textbf{(B) }21}</math>.
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~andliu766
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== Solution 3 (Same as solution 1 but Using 100=99+1)==
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<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.
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~woh123
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== Video Solution by Pi Academy ==
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https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/sEk9jQnMzfk
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~Thesmartgreekmathdude
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== Video Solution by FrankTutor ==
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https://youtu.be/g2RxRsxrp2Y
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== Video Solution 1 by Power Solve ==
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https://youtu.be/j-37jvqzhrg?si=rWQoAYu7QsZP8ty4&t=407
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=6SQ74nt3ynw
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==See also==
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{{AMC10 box|year=2024|ab=A|num-b=3|num-a=5}}
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{{AMC12 box|year=2024|ab=A|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 22:06, 30 November 2024

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 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

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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