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

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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(Make the problem easier)
+
==Solution 2 (making the problem easier)==
  
−
If you did not see the pattern.
+
If you did not see the pattern, then we may solve a easier problem.
  
−
then solve a easier problem
 
 
What is the product of all positive odd integers less than <math>10</math>?
 
What is the product of all positive odd integers less than <math>10</math>?
  
−
1(3)(5)(7)(9) = 945
+
1(3)(5)(7)(9) = 945.
−
we have
 
−
<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{(E)}\ \dfrac{5000!}{2^{5000}}</math>
 
  
−
but now we had
+
Originally, we had
−
<math>\text{(A)}\ \dfrac{10!}{(5!)^2}\qquad \text{(B)}\ \dfrac{10!}{2^{5}}\qquad
+
<math>\textbf{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad
−
\text{(C)}\ \dfrac{9!}{2^{5}}\qquad \text{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}\qquad
+
\textbf{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad
−
\text{(E)}\ \dfrac{5!}{2^{5}}</math>
+
\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  
 
which expression equals 945  
−
<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
+
<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 12:02, 5 November 2024

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

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 $10$?

1(3)(5)(7)(9) = 945.

Originally, we had $\textbf{(A)}\ \dfrac{10000!}{(5000!)^2}\qquad \text{(B)}\ \dfrac{10000!}{2^{5000}}\qquad \textbf{(C)}\ \dfrac{9999!}{2^{5000}}\qquad \text{(D)}\ \dfrac{10000!}{2^{5000} \cdot 5000!}\qquad \textbf{(E)}\ \dfrac{5000!}{2^{5000}}$

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

which expression equals 945

$\textbf{(A)}\ \dfrac{10!}{(5!)^2}$ = 252 way too small

$\textbf{(B)}\ \dfrac{10!}{2^{5}}$ = is way too big, 113400

$\textbf{(C)}\ \dfrac{9!}{2^{5}}$ = is just 113400 divided by 10(11340), so still too big

$\textbf{(D)}\ \dfrac{10!}{2^{5} \cdot 5!}$ = 113400/120= 945, just perfect

$\textbf{(E)}\ \dfrac{5!}{2^{5}}$ = 3.75 or just too small

So D is equal to 945, thus the answer is $\boxed{\textbf{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}}$.

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

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