AMC 8 2016 Problem 24

AMC8 2016 年第 24 题

整除与因数★★★★☆

The digits 1 , 2 , 3 , 4 , and 5 are each used once to write a five-digit number

PQRST

. The three-digit number

PQR

is divisible by 4 , the three-digit number

QRS

is divisible by 5 , and the three-digit number

RST

is divisible by 3 . What is

P

数字 1，2，3，4，5 各使用一次组成一个五位数 PQRST，其中三位数 PQR 能被 4 整除，三位数 QRS 能被 5 整除，三位数 RST 能被 3 整除。问 P 是多少？

正确答案 · Correct Answer

解析 · Solution

解法1

常规解法 · 逐条件锁定

\overline{QRS}\mid5\Rightarrow S=5

。

\overline{RST}\mid3\Rightarrow R+5+T\equiv0\pmod3\Rightarrow R+T\equiv1\pmod3

，余下数

\{1,2,3,4\}

里满足的 $(R,T)$ 有限。再用

\overline{PQR}\mid4

（末两位

\overline{QR}\mid4

）筛。逐一枚举得唯一排列 PQRST=14532，故 P=1。

Solution 1

Elementary · lock it down condition by condition Since

\overline{QRS}

is divisible by 5, S=5. From

\overline{RST}

divisible by 3,

R+5+T\equiv0\pmod3

, so

R+T\equiv1\pmod3

, which limits $(R,T)$ among the remaining

\{1,2,3,4\}

. Filtering with

\overline{PQR}

divisible by 4 (only

\overline{QR}

matters) leaves the unique arrangement PQRST=14532, so P=1.

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