AMC10 2024 A

AMC10 2024 A · Q18

AMC10 2024 A · Q18. It mainly tests Remainders & modular arithmetic, Base representation.

There are exactly $K$ positive integers $b$ with $5 \leq b \leq 2024$ such that the base-$b$ integer $2024_b$ is divisible by $16$ (where $16$ is in base ten). What is the sum of the digits of $K$?

存在恰好$K$个正整数$b$满足$5 \leq b \leq 2024$，使得$b$进制整数$2024_b$能被$16$（十进制）整除。求$K$的各位数字之和？

(A) 16 16

(B) 17 17

(D) 20 20

(E) 21 21

Answer

Correct choice: (D)

正确答案：（D）

Solution

$2024_b = 2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8$. If $b$ is even, then $b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8$. If $b$ is odd, then $b^2\equiv 1\pmod 8\implies b^3+b+2\equiv 2b+2\pmod 8$ so $2b+2\equiv 0\pmod 8\implies b+1\equiv 0\pmod 4\implies b\equiv 3,7\pmod 8$. Now $8\mid 2024$ so $\frac38\cdot 2024=759$ but one of the answers we got from that, $3$, is too small, so $759 - 1 = 758\implies\boxed{\textbf{(D) }20}$.

$2024_b = 2b^3 + 2b + 4 \equiv 0 \pmod{16}$，即$b^3 + b + 2 \equiv 0 \pmod{8}$。若$b$为偶数，则$b + 2 \equiv 0 \pmod{8}$，即$b \equiv 6 \pmod{8}$。若$b$为奇数，则$b^2 \equiv 1 \pmod{8}$，故$b^3 + b + 2 \equiv 2b + 2 \pmod{8}$，即$2b + 2 \equiv 0 \pmod{8}$，$b + 1 \equiv 0 \pmod{4}$，$b \equiv 3,7 \pmod{8}$。现在$8 \mid 2024$，故$\frac{3}{8} \cdot 2024 = 759$，但其中$3$太小，需减$1$，得$758 \implies \boxed{\textbf{(D) }20}$。

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