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

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~andliu766
 
~andliu766
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/sEk9jQnMzfk
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~Thesmartgreekmathdude
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=3|num-a=5}}
 
{{AMC10 box|year=2024|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:01, 8 November 2024

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

Video Solution by Daily Dose of Math

https://youtu.be/sEk9jQnMzfk

~Thesmartgreekmathdude

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

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