AMC8 2016

AMC8 2016 · Q20

AMC8 2016 · Q20. It mainly tests GCD & LCM.

The least common multiple of $a$ and $b$ is 12, and the least common multiple of $b$ and $c$ is 15. What is the least possible value of the least common multiple of $a$ and $c$?

$a$ 和 $b$ 的最小公倍数是 12，$b$ 和 $c$ 的最小公倍数是 15。$a$ 和 $c$ 的最小公倍数的最小可能值是多少？

(A) 20 20

(B) 30 30

(D) 120 120

(E) 180 180

Answer

Correct choice: (A)

正确答案：（A）

Solution

If b = 1, then a = 12 and c = 15, and the least common multiple of a and c is 60. If b > 1, then any prime factor of b must also be a factor of both 12 and 15, and thus the only possible value is b = 3. In this case, a must be a multiple of 4 and a divisor of 12, so a = 4 or a = 12. Similarly, c must be a multiple of 5 and a divisor of 15, so c = 5 or c = 15. It follows that the least common multiple of a and c must be a multiple of 20. When a = 4, b = 3, and c = 5, the least common multiple of a and c is exactly 20.

若 b = 1，则 a = 12 且 c = 15，此时 a 与 c 的最小公倍数为 60。若 b > 1，则 b 的任一质因数都必须同时整除 12 和 15，因此 b 只能取 3。在这种情况下，a 必须是 4 的倍数且为 12 的约数，所以 a = 4 或 a = 12。类似地，c 必须是 5 的倍数且为 15 的约数，所以 c = 5 或 c = 15。于是 a 与 c 的最小公倍数必须是 20 的倍数。当 a = 4、b = 3、c = 5 时，a 与 c 的最小公倍数恰为 20。

Topics

NT.003 GCD & LCM