2004 AMC 10B Problems/Problem 6

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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