Difference between revisions of "2001 AMC 12 Problems/Problem 5"
(→See Also) |
(→Solution) |
||
Line 28: | Line 28: | ||
\text{(C)}\ \dfrac{9!}{2^{5}}\qquad \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}\qquad | \text{(C)}\ \dfrac{9!}{2^{5}}\qquad \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}\qquad | ||
\text{(E)}\ \dfrac{5!}{2^{5}}</math> | \text{(E)}\ \dfrac{5!}{2^{5}}</math> | ||
+ | |||
+ | which expression equals 945 | ||
+ | $\text{(A)}\ \dfrac{10!}{(5!)^2}\qquad | ||
+ | |||
== 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 00:08, 11 March 2023
Problem
What is the product of all positive odd integers less than ?
Solution
Therefore the answer is .
Solution 2(Make the problem easier)
If you did not see the pattern.
then solve a easier problem
What is the product of all positive odd integers less than ?
1(3)(5)(7)(9) = 945
we have
but now we have
which expression equals 945 $\text{(A)}\ \dfrac{10!}{(5!)^2}\qquad
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.