Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 19: Line 19:
  
 
1(3)(5)(7)(9) = 945
 
1(3)(5)(7)(9) = 945
−
we have
+
we had
 
<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
 
\text{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad
 
\text{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad
 
\text{(E)}\ \dfrac{5000!}{2^{5000}}</math>
 
\text{(E)}\ \dfrac{5000!}{2^{5000}}</math>
  
−
but now we had
+
but now we have
 
<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

Revision as of 23:17, 10 March 2023

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

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 $10$?

1(3)(5)(7)(9) = 945 we had $\text{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad \text{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad \text{(E)}\ \dfrac{5000!}{2^{5000}}$

but now we have $\text{(A)}\ \dfrac{10!}{(5!)^2}\qquad \text{(B)}\ \dfrac{10!}{2^{5}}\qquad \text{(C)}\ \dfrac{9!}{2^{5}}\qquad \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}\qquad \text{(E)}\ \dfrac{5!}{2^{5}}$

which expression equals 945 $\text{(A)}\ \dfrac{10!}{(5!)^2}$ = 252 way too small $\text{(B)}\ \dfrac{10!}{2^{5}}$ = is way too big 113400 $\text{(C)}\ \dfrac{9!}{2^{5}}$ = is just 113400 divided by 10(11340), so still too big $\text{(D)}\ \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}$ = 113400/120= 945, just perfect $\text{(E)}\ \dfrac{5!}{2^{5}}$ = 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_5&oldid=190806"