2000 AMC 12 第 6 题

Question

从

4

和

18

之间选出两个不同的质数。用它们的乘积减去它们的和，可能得到下列哪个数？

Two different prime numbers between

4

and

18

are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

21

60

119

180

231

答案：C

知识点：质数奇偶性因式分解

难度评级：1310

小提示：

这些质数是

5, 7, 11, 13

和

17

；注意

xy - (x + y)

= (x - 1)(y - 1) - 1

The primes are

5, 7, 11, 13,

and

17

; note that

xy - (x + y)

= (x - 1)(y - 1) - 1

大提示：

两个质数都是奇数，所以

(x-1)(y-1)-1 \equiv 3 \pmod 4

；再检验剩下的候选项。

Both primes are odd, so

(x-1)(y-1)-1 \equiv 3 \pmod 4;

then test the remaining candidates

解答：

4

和

18

之间的质数是

5, 7, 11, 13

和

17

。

对这样的两个质数，

xy - (x + y)

= (x - 1)(y - 1) - 1

是两个偶数的乘积减去

1

，因此它同余于

3 \pmod 4

。这样只剩下

119

和

231

。后者要求

(x-1)(y-1)=232

，但

\{4,6,10,12,16\}

中没有两个不同的数的乘积等于

232

。

实际上，

\begin{aligned} 11 \cdot 13 - (11 + 13) &= 143 - 24 \\ &= 119 \end{aligned}\text{。}

因此，正确答案是 C。

The primes between

4

and

18

are

5, 7, 11, 13,

and

17.

For two such primes,

xy - (x + y)

= (x - 1)(y - 1) - 1

is a product of two even numbers minus

1,

hence it is

3 \pmod 4.

This leaves

119

and

231.

The latter would require

(x-1)(y-1)=232,

but no two distinct numbers in

\{4,6,10,12,16\}

have product

232.

Indeed,

\begin{aligned} 11 \cdot 13 - (11 + 13) &= 143 - 24 \\ &= 119. \end{aligned}

Thus, the correct answer is C.

2000 AMC 12 第 6 题

答案：C

难度评级：1310

小提示：

大提示：

解答：

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