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 $\boxed{\mathrm{(C) \ } 99! \cdot 100!}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2004 AMC 10B Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See also