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

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.

Page

Toolbox

Search

2024 AMC 10A Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

See also