1990 AIME 第 10 题

Question

集合

A=\{z:z^{18}=1\}

和

B=\{w:w^{48}=1\}

都是复单位根的集合。集合

C=\{zw:z\in A,\ w\in B\}

也是复单位根的集合。

C

中有多少个不同的元素？

The sets

A=\{z:z^{18}=1\}

and

B=\{w:w^{48}=1\}

are both sets of complex roots of unity. The set

C=\{zw:z\in A,\ w\in B\}

is also a set of complex roots of unity. How many distinct elements are in

C?

答案

将各个根写成指数形式，其辐角分别是

\frac{2\pi}{18}

和

\frac{2\pi}{48}

的整数倍

Write the roots as exponentials whose arguments are multiples of

\frac{2\pi}{18}

and

\frac{2\pi}{48}

这些辐角的和生成

\frac{2\pi}{\operatorname{lcm}(18,48)}

的所有整数倍

The sums of those arguments generate all multiples of

\frac{2\pi}{\operatorname{lcm}(18,48)}

C

中各乘积的辐角为

2\pi\left(\frac a{18}+\frac b{48}\right)=\frac{2\pi(8a+3b)}{144}\text{。}

由于

\gcd(8,3)=1

，

8a+3b

的剩余类可生成模

144

的每个剩余类。因此

C

恰好是

144

次单位根的集合。等价地，其阶为

\operatorname{lcm}(18,48)=144

。

The arguments of products in

C

are

2\pi\left(\frac a{18}+\frac b{48}\right)=\frac{2\pi(8a+3b)}{144}.

Since

\gcd(8,3)=1,

the residues

8a+3b

generate every residue modulo

144.

Thus

C

is precisely the set of

144

th roots of unity. Equivalently, its order is

\operatorname{lcm}(18,48)=144.

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