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Difference between revisions of "2004 AMC 10B Problems/Problem 6"
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<math> \mathrm{(A) \ } 98! \cdot 99! \qquad \mathrm{(B) \ } 98! \cdot 100! \qquad \mathrm{(C) \ } 99! \cdot 100! \qquad \mathrm{(D) \ } 99! \cdot 101! \qquad \mathrm{(E) \ } 100! \cdot 101! </math>
 
<math> \mathrm{(A) \ } 98! \cdot 99! \qquad \mathrm{(B) \ } 98! \cdot 100! \qquad \mathrm{(C) \ } 99! \cdot 100! \qquad \mathrm{(D) \ } 99! \cdot 101! \qquad \mathrm{(E) \ } 100! \cdot 101! </math>
  
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==Solution==
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==Solution 1==
  
 
Using the fact that <math>n! = n\cdot (n-1)!</math>, we can write:
 
Using the fact that <math>n! = n\cdot (n-1)!</math>, we can write:
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* <math>A=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = 9\cdot 11\cdot(98!)^2</math>
 
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* <math>B=100 \cdot 99 \cdot (98!)^2 = 9\cdot 11\cdot (10\cdot 98!)^2</math>
 
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* <math>C=100\cdot (99!)^2 = (10\cdot 99!)^2</math>
 
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* <math>D=101\cdot 100\cdot (99!)^2 = 101 \cdot(10\cdot 99!)^2</math>
 
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* <math>E=101\cdot (100!)^2</math>
 
  
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Clearly <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math> is a square, and as <math>9</math>, <math>11</math>, and <math>101</math> are primes, none of the other four are squares.
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<cmath>\begin{align} A&=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = 11\cdot3^2\cdot(98!)^2 \\ B&=100 \cdot 99 \cdot (98!)^2 = 11\cdot10^2\cdot3^2\cdot( 98!)^2 \\ C&=100\cdot (99!)^2 = 10^2\cdot (99!)^2\\ D&=101\cdot 100\cdot (99!)^2 = 101 \cdot 10^2 \cdot (99!)^2\\ E& =101\cdot (100!)^2 \end{align}</cmath>
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We see that <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math> is a square, and because <math>11</math>, and <math>101</math> are primes, none of the other four choices are squares.
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==Solution 2==
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Notice that <math>99! \cdot 99!</math> is a perfect square (obviously). Also, <math>100 = 10^2</math>. Thus,
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<cmath>(99!)^2 \cdot 10^2 = 99! \cdot 100!</cmath>
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is a perfect square, so the answer is <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:48, 26 July 2020

Problem

Which of the following numbers is a perfect square?

$\mathrm{(A) \ } 98! \cdot 99! \qquad \mathrm{(B) \ } 98! \cdot 100! \qquad \mathrm{(C) \ } 99! \cdot 100! \qquad \mathrm{(D) \ } 99! \cdot 101! \qquad \mathrm{(E) \ } 100! \cdot 101!$

Solution 1

Using the fact that $n! = n\cdot (n-1)!$, we can write:

\begin{align} A&=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = 11\cdot3^2\cdot(98!)^2 \\ B&=100 \cdot 99 \cdot (98!)^2 = 11\cdot10^2\cdot3^2\cdot( 98!)^2 \\ C&=100\cdot (99!)^2 = 10^2\cdot (99!)^2\\ D&=101\cdot 100\cdot (99!)^2 = 101 \cdot 10^2 \cdot (99!)^2\\ E& =101\cdot (100!)^2 \end{align}

We see that $\boxed{\mathrm{(C) \ } 99! \cdot 100!}$ is a square, and because $11$, and $101$ are primes, none of the other four choices are squares.


Solution 2

Notice that $99! \cdot 99!$ is a perfect square (obviously). Also, $100 = 10^2$. Thus, \[(99!)^2 \cdot 10^2 = 99! \cdot 100!\] is a perfect square, so the answer is $\boxed{\mathrm{(C) \ } 99! \cdot 100!}$.

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 10 Problems and Solutions

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