1975 AMC 12 第 29 题

Question

大于

(\sqrt3+\sqrt2)^6

的最小整数是多少？

What is the smallest integer larger than

(\sqrt3+\sqrt2)^6?

972

971

970

969

968

将该式与其共轭式

(\sqrt3-\sqrt2)^6

配对

Pair the expression with its conjugate

(\sqrt3-\sqrt2)^6

两式之和为整数，而共轭式的值严格介于

0

与

1

之间

Their sum is an integer, while the conjugate term lies strictly between

0

and

1

令

\alpha=\sqrt3+\sqrt2

、

\beta=\sqrt3-\sqrt2

。将它们的六次方相加，所有含奇次根式的项都会抵消：

\begin{aligned} \alpha^6+\beta^6 &=2\bigl(3^3+15(3^2)(2)\\ &\qquad+15(3)(2^2)+2^3\bigr)\\ &=970\text{。} \end{aligned}

因为

0\lt\beta\lt1

，所以

969\lt\alpha^6=970-\beta^6\lt970

。因此，大于它的最小整数为

970

。

因此，正确答案是 C。

Let

\alpha=\sqrt3+\sqrt2

and

\beta=\sqrt3-\sqrt2.

Adding their sixth powers cancels all odd radical terms:

\begin{aligned} \alpha^6+\beta^6 &=2\bigl(3^3+15(3^2)(2)\\ &\qquad+15(3)(2^2)+2^3\bigr)\\ &=970. \end{aligned}

Since

0\lt\beta\lt1,

we have

969\lt\alpha^6=970-\beta^6\lt970.

Thus the smallest larger integer is

970.

Therefore, the correct answer is C.

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