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：

其他年份的第 27 题