Difference between revisions of "2001 AMC 12 Problems/Problem 5"
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| − | == Problem == | + | ==Problem== |
What is the product of all positive odd integers less than 10000? | What is the product of all positive odd integers less than 10000? | ||
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\text{(E)}\ \dfrac{5000!}{2^{5000}}</math> | \text{(E)}\ \dfrac{5000!}{2^{5000}}</math> | ||
| − | == Solution == | + | ==Solution== |
<math>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!}</math> | <math>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!}</math> | ||
| − | <math>\text{(D)}</math> | + | Therefore the answer is <math>\boxed{\text{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}}</math>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=4|num-a=6}} | {{AMC12 box|year=2001|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:07, 29 March 2020
Problem
What is the product of all positive odd integers less than 10000?
Solution
Therefore the answer is 
.
See Also
| 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. 
