AMC12 2024 A

AMC12 2024 A · Q3

AMC12 2024 A · Q3. It mainly tests Arithmetic misc, GCD & LCM.

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?

数字 $2024$ 被写成若干（不一定不同）两位数的和。需要最少多少个两位数来表示这个和？

(A) 20 20

(B) 21 21

(D) 23 23

(E) 24 24

Answer

Correct choice: (B)

正确答案：（B）

Solution

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.

为了使两位数的个数最少，应最大化两位数，选择尽可能多的 $99$。因为 $2024=99\cdot20+44\cdot1$，所以选择二十个 $99$ 和一个 $44$，总共 $\boxed{\textbf{(B) }21}$ 个两位数。

Topics