AMC10 2018 A

AMC10 2018 A · Q22

AMC10 2018 A · Q22. It mainly tests Primes & prime factorization, GCD & LCM.

Let $a, b, c, d$ be positive integers such that $\gcd(a, b) = 24$, $\gcd(b, c) = 36$, $\gcd(c, d) = 54$, and $70 < \gcd(d, a) < 100$. Which of the following must be a divisor of $a$?

设 $a, b, c, d$ 为正整数，使得 $\gcd(a, b) = 24$，$\gcd(b, c) = 36$，$\gcd(c, d) = 54$，且 $70 < \gcd(d, a) < 100$。以下哪个必须是 $a$ 的因数？

(A) 5 5

(B) 7 7

(D) 13 13

(E) 17 17

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Because $\gcd(a,b)=24=2^3\cdot 3$ and $\gcd(b,c)=36=2^2\cdot 3^2$, it follows that $a$ is divisible by $2$ and $3$ but not by $3^2$. Similarly, because $\gcd(b,c)=2^2\cdot 3^2$ and $\gcd(c,d)=54=2\cdot 3^3$, it follows that $d$ is divisible by $2$ and $3$ but not by $2^2$. Therefore $\gcd(d,a)=2\cdot 3\cdot n$, where $n$ is a product of primes that do not include $2$ or $3$. Because $70<\gcd(d,a)<100$ and $n$ is an integer, it must be that $12\le n\le 16$, so $n=13$, and $13$ must also be a divisor of $a$. The conditions are satisfied if $a=2^3\cdot 3\cdot 13=312$, $b=2^3\cdot 3^2=72$, $c=2^2\cdot 3^3=108$, and $d=2\cdot 3^3\cdot 13=702$.

答案（D）：因为 $\gcd(a,b)=24=2^3\cdot 3$ 且 $\gcd(b,c)=36=2^2\cdot 3^2$，可知 $a$ 能被 $2$ 和 $3$ 整除，但不能被 $3^2$ 整除。类似地，因为 $\gcd(b,c)=2^2\cdot 3^2$ 且 $\gcd(c,d)=54=2\cdot 3^3$，可知 $d$ 能被 $2$ 和 $3$ 整除，但不能被 $2^2$ 整除。因此 $\gcd(d,a)=2\cdot 3\cdot n$，其中 $n$ 是不包含质因子 $2$ 或 $3$ 的若干质数的乘积。由于 $70<\gcd(d,a)<100$ 且 $n$ 为整数，必有 $12\le n\le 16$，所以 $n=13$，并且 $13$ 也必须是 $a$ 的因子。若取 $a=2^3\cdot 3\cdot 13=312$，$b=2^3\cdot 3^2=72$，$c=2^2\cdot 3^3=108$，$d=2\cdot 3^3\cdot 13=702$，则满足条件。

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