Difference between revisions of "2001 AMC 12 Problems/Problem 5"

Revision as of 18:07, 29 March 2020

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

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

@@ Line 1: / Line 1: @@
 ==Problem==
-What is the product of all positive odd integers less than 10000?
+What is the product of all positive odd integers less than <math>10000</math>?
 <math>\text{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad

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Difference between revisions of "2001 AMC 12 Problems/Problem 5"

Revision as of 18:07, 29 March 2020

Problem

Solution

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