AMC12 2024 A

AMC12 2024 A · Q6

AMC12 2024 A · Q6. It mainly tests Primes & prime factorization, Number theory misc.

The product of three integers is $60$. What is the least possible positive sum of the three integers?

三个整数的乘积是$60$。这三个整数的最小正和是多少？

(A) 2 2

(B) 3 3

(D) 6 6

(E) 13 13

Answer

Correct choice: (B)

正确答案：（B）

Solution

We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split $60$ into three factors and choose negativity. We notice that $10\cdot6\cdot1=10\cdot(-6)\cdot(-1)=60$, and trying other combinations does not yield lesser results so the answer is $10-6-1=\boxed{\textbf{(B) }3}$.

我们注意到最优解涉及两个负数和一个正数。因此我们可以将$60$分解成三个因数并选择符号。我们注意到$10\cdot6\cdot1=10\cdot(-6)\cdot(-1)=60$，尝试其他组合不会得到更小的结果，所以答案是$10-6-1=\boxed{\textbf{(B) }3}$。

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