2001 AMC 12 Problem 5

Question

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}}

The product of all odd numbers equals

10000!

divided by the product of all even numbers up to

10000

The even numbers are

2 \cdot 4 \cdots 10000

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

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.

Problem 5 in other years