Difference between revisions of "2001 AMC 12 Problems/Problem 5"
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\text{(E)}\ \dfrac{5000!}{2^{5000}}</math> | \text{(E)}\ \dfrac{5000!}{2^{5000}}</math> | ||
− | but now we | + | but now we had |
<math>\text{(A)}\ \dfrac{10!}{(5!)^2}\qquad \text{(B)}\ \dfrac{10!}{2^{5}}\qquad | <math>\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{(C)}\ \dfrac{9!}{2^{5}}\qquad \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}\qquad | ||
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which expression equals 945 | which expression equals 945 | ||
− | <math>\text{(A)}\ \dfrac{10!}{(5!)^2}</math> = | + | <math>\text{(A)}\ \dfrac{10!}{(5!)^2}</math> = 252 way too small |
+ | <math>\text{(B)}\ \dfrac{10!}{2^{5}}</math> = is way too big 113400 | ||
+ | <math>\text{(C)}\ \dfrac{9!}{2^{5}}</math> = is just 113400 divided by 10(11340), so still too big | ||
+ | <math>\text{(D)}\ \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}</math> = 113400/120= 945, just perfect | ||
+ | <math>\text{(E)}\ \dfrac{5!}{2^{5}}</math> = 3.75 or just too small | ||
+ | |||
+ | So D is equal to 945, thus the answer is D | ||
== 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 23:16, 10 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 had
which expression equals 945 = 252 way too small = is way too big 113400 = is just 113400 divided by 10(11340), so still too big = 113400/120= 945, just perfect = 3.75 or just too small
So D is equal to 945, thus the answer is D
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.