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

Revision as of 02:29, 31 January 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

4. Can you identify the fatty? https://www.linkedin.com/in/manuel-norse-621058109/

https://www.oregonmathcircle.org/ (look at the bald headahh diddler)

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

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 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

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. Error creating thumbnail: File missing

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