AMC12 2024 A

AMC12 2024 A · Q11

AMC12 2024 A · Q11. 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 $5 \leq b \leq 2024$，使得基数 $b$ 的整数 $2024_b$ 能被 $16$（十进制）整除。$K$ 的各位数字之和是多少？

(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 $759 - 1 = 758\implies\boxed{\textbf{(D) }20}$.

$2024_b = 2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8$。若 $b$ 为偶数，则 $b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8$。若 $b$ 为奇数，则 $b^2\equiv 1\pmod 8\implies b^3+b+2\equiv 2b+2\pmod 8$，所以 $2b+2\equiv 0\pmod 8\implies b+1\equiv 0\pmod 4\implies b\equiv 3,7\pmod 8$。现在 $8\mid 2024$ 所以 $\frac38\cdot 2024=759$，但其中一个解 $3$ 太小，所以 $759 - 1 = 758\implies\boxed{\textbf{(D) }20}$。

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