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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) ...

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 (Problems • 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
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