Difference between revisions of "2004 AMC 10B Problems/Problem 6"

Revision as of 12:12, 4 July 2013

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.

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

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Revision as of 13:32, 7 February 2009 (view source) Misof (talk \| contribs) (New page: ==Problem== Which of the following numbers is a perfect square? <math> \mathrm{(A) \ } 98! \cdot 99! \qquad \mathrm{(B) \ } 98! \cdot 100! \qquad \mathrm{(C) \ } 99! \cdot 100! \qquad \m...)		Revision as of 12:12, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	{{AMC10 box\|year=2004\|ab=B\|num-b=5\|num-a=7}}		{{AMC10 box\|year=2004\|ab=B\|num-b=5\|num-a=7}}
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Difference between revisions of "2004 AMC 10B Problems/Problem 6"

Revision as of 12:12, 4 July 2013

Problem

Solution

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