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 <math>10000</math>? |
<math>\text{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad | <math>\text{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad | ||
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<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> | ||
| − | Therefore the answer is <math>\boxed{\text{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}}</math>. | + | Therefore the answer is <math>\boxed{\textbf{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}}</math>. |
| + | |||
| + | ==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 <math>10</math>? | ||
| + | |||
| + | 1(3)(5)(7)(9) = 945. | ||
| + | |||
| + | Originally, we had | ||
| + | <math>\textbf{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad | ||
| + | \textbf{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad | ||
| + | \textbf{(E)}\ \dfrac{5000!}{2^{5000}}</math> | ||
| + | |||
| + | but now we have | ||
| + | <math>\textbf{(A)}\ \dfrac{10!}{(5!)^2}\qquad \text{(B)}\ \dfrac{10!}{2^{5}}\qquad | ||
| + | \textbf{(C)}\ \dfrac{9!}{2^{5}}\qquad \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}\qquad | ||
| + | \textbf{(E)}\ \dfrac{5!}{2^{5}}</math> | ||
| + | |||
| + | which expression equals 945 | ||
| + | |||
| + | <math>\textbf{(A)}\ \dfrac{10!}{(5!)^2}</math> = 252 way too small | ||
| + | |||
| + | <math>\textbf{(B)}\ \dfrac{10!}{2^{5}}</math> = is way too big, 113400 | ||
| + | |||
| + | <math>\textbf{(C)}\ \dfrac{9!}{2^{5}}</math> = is just 113400 divided by 10(11340), so still too big | ||
| + | |||
| + | <math>\textbf{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}</math> = 113400/120= 945, just perfect | ||
| + | |||
| + | <math>\textbf{(E)}\ \dfrac{5!}{2^{5}}</math> = 3.75 or just too small | ||
| + | |||
| + | So D is equal to 945, thus the answer is <math>\boxed{\textbf{(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}} | ||
Latest revision as of 23:30, 29 December 2023
Problem
What is the product of all positive odd integers less than 
?
Solution
Therefore the answer is 
.
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 
?
1(3)(5)(7)(9) = 945.
Originally, we had
but now we have
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 .
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. 
