AMC10 2022 A

AMC10 2022 A · Q7

AMC10 2022 A · Q7. It mainly tests Primes & prime factorization, GCD & LCM.

The least common multiple of a positive integer $n$ and $18$ is $180$, and the greatest common divisor of $n$ and $45$ is $15$. What is the sum of the digits of $n$?

一个正整数 $n$ 与 $18$ 的最小公倍数是 $180$，$n$ 与 $45$ 的最大公因数是 $15$。$n$ 的各位数字之和是多少？

(A) 3 3

(B) 6 6

(D) 9 9

(E) 12 12

Answer

Correct choice: (B)

正确答案：（B）

Solution

Note that \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that: 1. From the least common multiple condition, we have \[\operatorname{lcm}(n,18) = \operatorname{lcm}(2^a\cdot3^b\cdot5^c,2\cdot3^2) = 2^{\max(a,1)}\cdot3^{\max(b,2)}\cdot5^{\max(c,0)} = 2^2\cdot3^2\cdot5,\] from which $a=2, b\in\{0,1,2\},$ and $c=1.$ 2. From the greatest common divisor condition, we have \[\gcd(n,45) = \gcd(2^2\cdot3^b\cdot5,3^2\cdot5) = 2^{\min(2,0)}\cdot3^{\min(b,2)}\cdot5^{\min(1,1)} = 3\cdot5,\] from which $b=1.$ Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{\textbf{(B) } 6}.$

注意 \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} 设 $n = 2^a\cdot3^b\cdot5^c$。由此得出： 1. 从最小公倍数条件，有 \[\operatorname{lcm}(n,18) = \operatorname{lcm}(2^a\cdot3^b\cdot5^c,2\cdot3^2) = 2^{\max(a,1)}\cdot3^{\max(b,2)}\cdot5^{\max(c,0)} = 2^2\cdot3^2\cdot5,\] 由此 $a=2, b\in\{0,1,2\},$ 和 $c=1$。 2. 从最大公因数条件，有 \[\gcd(n,45) = \gcd(2^2\cdot3^b\cdot5,3^2\cdot5) = 2^{\min(2,0)}\cdot3^{\min(b,2)}\cdot5^{\min(1,1)} = 3\cdot5,\] 由此 $b=1$。综合得出 $n=2^2\cdot3\cdot5=60$。其各位数字之和是 $6+0=\boxed{\textbf{(B) } 6}$。

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