AMC 8 2001 Problem 25
AMC8 2001 年第 25 题
整除与因数★★★★☆
There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
2、4、5 和 7 这四个数字每个恰好使用一次，一共可以组成 24 个四位整数。这 24 个四位整数中，只有一个是另一个的倍数。则这个数是下列哪一个？
- A.5724
- B.7245
- C.7254
- D.7425
- E.7542
正确答案 · Correct Answer
D
解析 · Solution
解法1
常规解法 · 试除
只在 5 个选项里试，用 2、3 去除并看商是否仍是 的排列：，数字 2、4、7、5 正是该集合，故 7425=3×2475，选 D。
巧解 · 缩小范围
倍数与被乘数都用同一组数字 ，商只能是 2 或 3（再大首位就变了）。试 恰是同集合排列，立得 D。
Solution 1
Elementary · trial division
Just test the five choices, dividing by small factors and checking the quotient still uses : , whose digits are 2, 4, 7, 5 — exactly the same set. So 7425=3×2475, choice D.
Slick · narrow the range
Both the multiple and its factor use the same digits , so the ratio can only be 2 or 3 (larger would change the leading digit). Testing, is again a permutation of the set, giving D.
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