1978 AMC 12 第 27 题

Question

大于

1

且满足下列条件的整数不止一个：用任意满足

2\le k\le11

的整数

k

除它，余数都是

1

。这样的整数中最小的两个相差多少？

There is more than one integer greater than

1

which, when divided by any integer

k

such that

2\le k\le11,

has a remainder of

1.

What is the difference between the two smallest such integers?

2310

2311

27{,}720

27{,}721

以上都不是

none of these

答案：C

知识点：最小公倍数模运算

难度评级：1780

小提示：

每个所求整数都比从

2

到

11

的所有整数的一个公倍数大

1

Each desired integer is

1

more than a common multiple of every integer from

2

through

11

大提示：

相邻两个这样的整数之差就是这些除数的最小公倍数

The difference of consecutive such integers is the least common multiple of those divisors

解答：

符合条件的整数对从

2

到

11

的每个整数取模都与

1

同余，因此也可将模数取为

\begin{aligned} L&=\operatorname{lcm}(2,\ldots,11)\\ &=2^3\cdot3^2\cdot5\cdot7\cdot11\\ &=27720\text{。} \end{aligned}

大于

1

的最小两个符合条件的整数是

L+1

和

2L+1

，它们的差为

L=27720

。

因此，正确答案是 C。

A qualifying integer is congruent to

1

modulo every integer from

2

through

11,

hence modulo

\begin{aligned} L&=\operatorname{lcm}(2,\ldots,11)\\ &=2^3\cdot3^2\cdot5\cdot7\cdot11\\ &=27720. \end{aligned}

The two smallest qualifying integers greater than

1

are

L+1

and

2L+1,

whose difference is

L=27720.

Therefore, the correct answer is C.

1978 AMC 12 第 27 题

答案：C

难度评级：1780

小提示：

大提示：

解答：

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