# 2001 AMC 12 Problems/Problem 5

## Problem

What is the product of all positive odd integers less than $10000$? $\text{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad \text{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad \text{(E)}\ \dfrac{5000!}{2^{5000}}$

## Solution $1 \cdot 3 \cdot 5 \cdots 9999 = \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdots 10000}{2 \cdot 4 \cdot 6 \cdots 10000}= \dfrac{10000!}{2^{5000} \cdot 1 \cdot 2 \cdot 3 \cdots 5000}= \dfrac{10000!}{2^{5000}\cdot5000!}$

Therefore the answer is $\boxed{\text{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}}$.

## Solution 2(making the problem easier)

If you did not see the pattern, then we may solve a easier problem.

What is the product of all positive odd integers less than $10$?

1(3)(5)(7)(9) = 945.

Originally, we had $\text{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad \text{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad \text{(E)}\ \dfrac{5000!}{2^{5000}}$

but now we have $\text{(A)}\ \dfrac{10!}{(5!)^2}\qquad \text{(B)}\ \dfrac{10!}{2^{5}}\qquad \text{(C)}\ \dfrac{9!}{2^{5}}\qquad \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}\qquad \text{(E)}\ \dfrac{5!}{2^{5}}$

which expression equals 945 $\text{(A)}\ \dfrac{10!}{(5!)^2}$ = 252 way too small $\text{(B)}\ \dfrac{10!}{2^{5}}$ = is way too big, 113400 $\text{(C)}\ \dfrac{9!}{2^{5}}$ = is just 113400 divided by 10(11340), so still too big $\text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}$ = 113400/120= 945, just perfect $\text{(E)}\ \dfrac{5!}{2^{5}}$ = 3.75 or just too small

So D is equal to 945, thus the answer is D

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 