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]]}} | ||
== Problem == | == Problem == | ||
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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> | <math>\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24</math> | ||
| − | == Solution == | + | == Solution 1 == |
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. | 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. | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | == Solution 2 == | ||
| + | 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>. | ||
| + | |||
| + | ~andliu766 | ||
| + | |||
| + | == 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 | ||
| + | |||
| + | == 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 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== | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=3|num-a=5}} | ||
| + | {{AMC12 box|year=2024|ab=A|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:53, 13 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.
Contents
Problem
The number 
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?
Solution 1
Since we want the least number of two-digit numbers, we maximize the two-digit numbers by choosing as many 
s as possible. Since 
we choose twenty s and one 
for a total of 
two-digit numbers.
~MRENTHUSIASM
Solution 2
We claim the answer is 
. This can be achieved by adding twenty 's and a 
. To prove that the answer cannot be less than or equal to 
, we note that the maximum value of the sum of or less two digit numbers is 
, which is smaller than , so we are done. Thus, the answer is .
~andliu766
Solution 3 (Same as solution 1 but Using 100=99+1)
. Since 
, 
. Therefore a total of two-digit numbers are needed.
~woh123
Video Solution by Pi Academy
https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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. 
