Difference between revisions of "2024 AMC 10A Problems/Problem 4"
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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 | ||
+ | |||
==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 16:06, 8 November 2024
Contents
[hide]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 |
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