AMC8 2001

AMC8 2001 · Q25

AMC8 2001 · Q25. It mainly tests Permutations, Divisibility & factors.

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?

有24个四位数，它们恰好各使用一次数字2、4、5、7。其中只有一个四位数是另一个的倍数。以下哪个是它？

(A) 5724 5724

(B) 7245 7245

(D) 7425 7425

(E) 7542 7542

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Six of the 24 numbers are in the 2000s, six in the 4000s, six in the 5000s and six in the 7000s. Doubling and tripling numbers in the 2000s produce possible solutions, but any multiple of those in the other sets is larger than 8000. Units digits of the numbers are 2, 4, 5 and 7, so their doubles will end in 4, 8, 0 and 4, respectively. Choice (A) 5724 ends in 4 but $5724/2=2862$, not one of the 24 numbers. Likewise, choice (C) 7254 produces $7254/2=3627$, also not one of the numbers. When the units digits are tripled the resulting units digits are 6, 2, 5 and 1 and choices (B) 7245, (D) 7425 and (E) 7542 are possibilities. Division by 3 yields 2415, 2475 and 2514 respectively. Only the second of these numbers is one of the 24 given numbers. Choice (D) is correct.

（D）这 24 个数中，有 6 个在 2000 段，6 个在 4000 段，6 个在 5000 段，6 个在 7000 段。将 2000 段中的数加倍或三倍会产生可能的解，但对其他几组中的数取倍数都会大于 8000。这些数的个位数字是 2、4、5 和 7，因此它们的两倍的个位分别是 4、8、0 和 4。选项（A）5724 的个位是 4，但 $5724/2=2862$，不是这 24 个数之一。同样，选项（C）7254 有 $7254/2=3627$，也不是这些数之一。将个位数字乘以 3 后得到的个位数字是 6、2、5 和 1，因此选项（B）7245、（D）7425 和（E）7542 都有可能。分别除以 3 得到 2415、2475 和 2514。只有第二个数是所给 24 个数之一。因此选项（D）正确。

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