Difference between revisions of "2024 AMC 10A Problems/Problem 4"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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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> | 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 | ~andliu766 | ||
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==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 16:08, 8 November 2024
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
