AMC12 2023 B

AMC12 2023 B · Q24

AMC12 2023 B · Q24. It mainly tests Primes & prime factorization, GCD & LCM.

Suppose that $a$, $b$, $c$ and $d$ are positive integers satisfying all of the following relations. \[abcd=2^6\cdot 3^9\cdot 5^7\] \[\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3\] \[\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2\] \[\text{lcm}(b,d)=2^2\cdot 3^3\cdot 5^2\] \[\text{lcm}(c,d)=2^2\cdot 3^3\cdot 5^2\] What is $\text{gcd}(a,b,c,d)$?

假设 $a$，$b$，$c$ 和 $d$ 是满足以下所有关系的正整数。 \[abcd=2^6\cdot 3^9\cdot 5^7\] \[\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3\] \[\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2\] \[\text{lcm}(b,d)=2^2\cdot 3^3\cdot 5^2\] \[\text{lcm}(c,d)=2^2\cdot 3^3\cdot 5^2\] $\text{gcd}(a,b,c,d)$ 是多少？

(A) 30 30

(B) 45 45

(D) 15 15

(E) 6 6

Answer

Correct choice: (C)

正确答案：（C）

Solution

Denote by $\nu_p (x)$ the number of prime factor $p$ in number $x$. We index Equations given in this problem from (1) to (7). First, we compute $\nu_2 (x)$ for $x \in \left\{ a, b, c, d \right\}$. Equation (5) implies $\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1$. Equation (2) implies $\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3$. Equation (6) implies $\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2$. Equation (1) implies $\nu_2 (a) + \nu_2 (b) + \nu_2 (c) + \nu_2 (d) = 6$. Therefore, all above jointly imply $\nu_2 (a) = 3$, $\nu_2 (d) = 2$, and $\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)$ or $\left( 1, 0 \right)$. Second, we compute $\nu_3 (x)$ for $x \in \left\{ a, b, c, d \right\}$. Equation (2) implies $\max \left\{ \nu_3 (a), \nu_3 (b) \right\} = 2$. Equation (3) implies $\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3$. Equation (4) implies $\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3$. Equation (1) implies $\nu_3 (a) + \nu_3 (b) + \nu_3 (c) + \nu_3 (d) = 9$. Therefore, all above jointly imply $\nu_3 (c) = 3$, $\nu_3 (d) = 3$, and $\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)$ or $\left( 2, 1 \right)$. Third, we compute $\nu_5 (x)$ for $x \in \left\{ a, b, c, d \right\}$. Equation (5) implies $\max \left\{ \nu_5 (b), \nu_5 (c) \right\} = 2$. Equation (2) implies $\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3$. Thus, $\nu_5 (a) = 3$. From Equations (5)-(7), we have either $\nu_5 (b) \leq 1$ and $\nu_5 (c) = \nu_5 (d) = 2$, or $\nu_5 (b) = 2$ and $\max \left\{ \nu_5 (c), \nu_5 (d) \right\} = 2$. Equation (1) implies $\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7$. Thus, for $\nu_5 (b)$, $\nu_5 (c)$, $\nu_5 (d)$, there must be two 2s and one 0. Therefore, \begin{align*} {\rm gcd} (a,b,c,d) & = \Pi_{p \in \{ 2, 3, 5\}} p^{\min\{ \nu_p (a), \nu_p(b) , \nu_p (c), \nu_p(d) \}} \\ & = 2^0 \cdot 3^1 \cdot 5^0 \\ & = \boxed{\textbf{(C) 3}}. \end{align*}

用 $\nu_p (x)$ 表示数 $x$ 中质因数 $p$ 的个数。将题目方程编号为 (1) 到 (7)。首先，计算 $x \in \left\{ a, b, c, d \right\}$ 的 $\nu_2 (x)$。方程 (5) 意味着 $\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1$。方程 (2) 意味着 $\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3$。方程 (6) 意味着 $\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2$。方程 (1) 意味着 $\nu_2 (a) + \nu_2 (b) + \nu_2 (c) + \nu_2 (d) = 6$。因此，所有以上联合意味着 $\nu_2 (a) = 3$，$\nu_2 (d) = 2$，且 $\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)$ 或 $\left( 1, 0 \right)$。其次，计算 $x \in \left\{ a, b, c, d \right\}$ 的 $\nu_3 (x)$。方程 (2) 意味着 $\max \left\{ \nu_3 (a), \nu_3 (b) \right\} = 2$。方程 (3) 意味着 $\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3$。方程 (4) 意味着 $\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3$。方程 (1) 意味着 $\nu_3 (a) + \nu_3 (b) + \nu_3 (c) + \nu_3 (d) = 9$。因此，所有以上联合意味着 $\nu_3 (c) = 3$，$\nu_3 (d) = 3$，且 $\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)$ 或 $\left( 2, 1 \right)$。第三，计算 $x \in \left\{ a, b, c, d \right\}$ 的 $\nu_5 (x)$。方程 (5) 意味着 $\max \left\{ \nu_5 (b), \nu_5 (c) \right\} = 2$。方程 (2) 意味着 $\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3$。因此，$\nu_5 (a) = 3$。从方程 (5)-(7)，我们有要么 $\nu_5 (b) \leq 1$ 且 $\nu_5 (c) = \nu_5 (d) = 2$，要么 $\nu_5 (b) = 2$ 且 $\max \left\{ \nu_5 (c), \nu_5 (d) \right\} = 2$。方程 (1) 意味着 $\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7$。因此，对于 $\nu_5 (b)$，$\nu_5 (c)$，$\nu_5 (d)$，必须有两个 $2$ 和一个 $0$。因此， \begin{align*} {\rm gcd} (a,b,c,d) & = \Pi_{p \in \{ 2, 3, 5\}} p^{\min\{ \nu_p (a), \nu_p(b) , \nu_p (c), \nu_p(d) \}} \\ & = 2^0 \cdot 3^1 \cdot 5^0 \\ & = \boxed{\textbf{(C) 3}}. \end{align*}

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