2004 AMC 10B Problems/Problem 6

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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

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

  • $A=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = 9\cdot 11\cdot(98!)^2$
  • $B=100 \cdot 99 \cdot (98!)^2 = 9\cdot 11\cdot (10\cdot 98!)^2$
  • $C=100\cdot (99!)^2 = (10\cdot 99!)^2$
  • $D=101\cdot 100\cdot (99!)^2 = 101 \cdot(10\cdot 99!)^2$
  • $E=101\cdot (100!)^2$

Clearly $\boxed{\mathrm{(C) \ } 99! \cdot 100!}$ is a square, and as $9$, $11$, and $101$ are primes, none of the other four are squares.

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