Difference between revisions of "2004 AMC 10B Problems/Problem 6"
(→Solution) |
m (→Solution) |
||
| Line 9: | Line 9: | ||
Using the fact that <math>n! = n\cdot (n-1)!</math>, we can write: | Using the fact that <math>n! = n\cdot (n-1)!</math>, we can write: | ||
| − | + | <cmath>\begin{aligned} 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{aligned}</cmath> | |
| − | + | We see that <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math> is a square, and because <math>11</math>, and <math>101</math> are primes, none of the other four choices are squares. | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
== See also == | == See also == | ||
Revision as of 13:49, 20 January 2014
Problem
Which of the following numbers is a perfect square?
Solution
Using the fact that 
, we can write:
\begin{aligned} 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{aligned} (Error compiling LaTeX. Unknown error_msg)
We see that 
is a square, and because 
, and 
are primes, none of the other four choices are squares.
See also
| 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. 
