Difference between revisions of "2024 AMC 10A Problems/Problem 4"
(I provided a third solution) |
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~andliu766 | ~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 == | + | == Video Solution by Daily Dose of Math == |
https://youtu.be/sEk9jQnMzfk | https://youtu.be/sEk9jQnMzfk | ||
Revision as of 08:08, 10 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
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. 
