1970 AMC 12 第 4 题

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：

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