Question

设 $S$ 为所有可表示成三个连续整数的平方和的数所组成的集合。则有：

Let $S$ be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that:

$S$ 中没有元素能被 $2$ 整除

No member of $S$ is divisible by $2$

$S$ 中没有元素能被 $3$ 整除，但有元素能被 $11$ 整除

No member of $S$ is divisible by $3$ but some member is divisible by $11$

$S$ 中没有元素能被 $3$ 或 $5$ 整除

No member of $S$ is divisible by $3$ or by $5$

$S$ 中没有元素能被 $3$ 或 $7$ 整除

No member of $S$ is divisible by $3$ or by $7$

以上都不对

None of these

答案：B

知识点：模运算整除性代数变形

难度评级：1760

小提示：

将三个整数写成 $n-1,n,n+1$

Write the integers as $n-1,n,n+1$

大提示：

它们的平方和是 $3n^2+2$ ；先对模 $3$ 考察，再检验一个模 $11$ 的余数

Their squared sum is $3n^2+2;$ examine it modulo $3$ and test a residue modulo $11$

解答：

这个和为 $\begin{gathered} (n-1)^2+n^2+(n+1)^2\\ =3n^2+2 \end{gathered}\text{。}$ 它总是同余于 $2\pmod3$ ，所以 $S$ 中没有元素能被 $3$ 整除。另一方面，取 $n=5$ 时得到 $3(25)+2=77$ ，它能被 $11$ 整除。

因此，正确答案是 B。

The sum is $\begin{gathered} (n-1)^2+n^2+(n+1)^2\\ =3n^2+2. \end{gathered}$ It is always congruent to $2\pmod3,$ so no member of $S$ is divisible by $3.$ On the other hand, taking $n=5$ gives $3(25)+2=77,$ which is divisible by $11.$

Therefore, the correct answer is B.

1970 AMC 12 第 4 题

答案：B

难度评级：1760

小提示：

大提示：

解答：

其他年份的第 4 题