AMC12 2000 A

AMC12 2000 A · Q1

AMC12 2000 A · Q1. It mainly tests Primes & prime factorization, Counting divisors.

In the year $2001$, the United States will host the International Mathematical Olympiad. Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$. What is the largest possible value of the sum $I + M + O$?

在 $2001$ 年，美国将举办国际数学奥林匹克竞赛。设 $I,M,$ 和 $O$ 是互不相同的正整数，使得乘积 $I \cdot M \cdot O = 2001$。求和 $I + M + O$ 的最大可能值。

(A) 23 23

(B) 55 55

(D) 111 111

(E) 671 671

Answer

Correct choice: (E)

正确答案：（E）

Solution

First, we need to recognize that a number is going to be largest only if, of the $3$ factors, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$. It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be $18$. Now, we use this process on $2001$ to get $667 * 3 * 1$ as our $3$ factors. Hence, we have $667 + 3 + 1 = \boxed{\text{(E) 671.}}$

首先，我们需要认识到：一个数要尽可能大，通常是因为这 $3$ 个因数中有两个很小。若要确认这一点，我们可以用一个较小的数来测试，比如 $30$。这样会更清楚地看到这一点，在这种情况下，$I + M + O$ 的值会是 $18$。现在，我们对 $2001$ 采用同样的过程，得到 $667 * 3 * 1$ 作为我们的 $3$ 个因数。因此，$667 + 3 + 1 = \boxed{\text{(E) 671.}}$

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