2001 AMC 12 第 5 题

Question

所有小于

10{,}000

的正奇整数的乘积是多少？

What is the product of all positive odd integers less than

10{,}000?

\dfrac{10000!}{(5000!)^2}

\dfrac{10000!}{2^{5000}}

\dfrac{9999!}{2^{5000}}

\dfrac{10000!}{2^{5000} \cdot 5000!}

\dfrac{5000!}{2^{5000}}

所有奇数的乘积等于

10000!

除以所有不超过

10000

的偶数的乘积

The product of all odd numbers equals

10000!

divided by the product of all even numbers up to

10000

这些偶数为

2 \cdot 4 \cdots 10000

= 2^{5000}(1 \cdot 2 \cdots 5000)

The even numbers are

2 \cdot 4 \cdots 10000

= 2^{5000}(1 \cdot 2 \cdots 5000)

从

1

到

10000

的所有整数的乘积是

10000!

，所以所有奇数的乘积等于

10000!

除以所有偶数的乘积。

偶数的乘积可分解为

\begin{gathered} 2 \cdot 4 \cdots 10000 \\ = 2^{5000}(1 \cdot 2 \cdots 5000) \\ = 2^{5000} \cdot 5000! \end{gathered}\text{。}

因此奇整数的乘积是

\dfrac{10000!}{2^{5000} \cdot 5000!}\text{。}

因此，正确答案是 D。

The product of every integer from

1

to

10000

is

10000!,

so the product of the odd ones is

10000!

divided by the product of the even ones.

The even numbers factor as

\begin{gathered} 2 \cdot 4 \cdots 10000 \\ = 2^{5000}(1 \cdot 2 \cdots 5000) \\ = 2^{5000} \cdot 5000!. \end{gathered}

Therefore the product of the odd integers is

\dfrac{10000!}{2^{5000} \cdot 5000!}.

Thus, the correct answer is D.

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