AMC10 2015 A

AMC10 2015 A · Q18

AMC10 2015 A · Q18. It mainly tests Basic counting (rules of product/sum), Base representation.

Hexadecimal (base-16) numbers are written using the numeric digits 0 through 9 as well as the letters $A$ through $F$ to represent 10 through 15. Among the first 1000 positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$?

十六进制（以 16 为底）数使用数字 0 到 9 以及字母 $A$ 到 $F$ 来表示 10 到 15。在前 1000 个正整数中，有 $n$ 个数的十六进制表示只包含数字（不含字母）。求 $n$ 的各位数字之和。

(A) 17 17

(B) 18 18

(D) 20 20

(E) 21 21

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Because $1000 = 3\cdot 16^2 + 14\cdot 16 + 8$, the largest number less than $1000$ whose hexadecimal representation contains only numeric digits is $3\cdot 16^2 + 9\cdot 16 + 9$. Thus the number of such positive integers is $n = 4\cdot 10\cdot 10 - 1 = 399$ ($0\cdot 16^2 + 0\cdot 16 + 0 = 0$ is excluded), and the sum of the digits of $n$ is $21$.

答案（E）：因为$1000 = 3\cdot 16^2 + 14\cdot 16 + 8$，小于$1000$且其十六进制表示只包含数字字符（0–9）的最大数是$3\cdot 16^2 + 9\cdot 16 + 9$。因此，这样的正整数个数为$n = 4\cdot 10\cdot 10 - 1 = 399$（排除$0\cdot 16^2 + 0\cdot 16 + 0 = 0$），并且$n$的各位数字之和为$21$。

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