Difference between revisions of "2004 AMC 10B Problems/Problem 6"
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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: | ||
| − | * <math>A=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = | + | * <math>A=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = 11\cdot(3\cdot98!)^2</math> |
| − | * <math>B=100 \cdot 99 \cdot (98!)^2 = | + | * <math>B=100 \cdot 99 \cdot (98!)^2 = 11\cdot (30\cdot 98!)^2 </math> |
* <math>C=100\cdot (99!)^2 = (10\cdot 99!)^2</math> | * <math>C=100\cdot (99!)^2 = (10\cdot 99!)^2</math> | ||
* <math>D=101\cdot 100\cdot (99!)^2 = 101 \cdot(10\cdot 99!)^2</math> | * <math>D=101\cdot 100\cdot (99!)^2 = 101 \cdot(10\cdot 99!)^2</math> | ||
* <math>E=101\cdot (100!)^2</math> | * <math>E=101\cdot (100!)^2</math> | ||
| − | Clearly <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math> is a square, and as | + | Clearly <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math> is a square, and as <math>11</math>, and <math>101</math> are primes, none of the other four are squares. |
== See also == | == See also == | ||
Revision as of 12:45, 20 January 2014
Problem
Which of the following numbers is a perfect square?
Solution
Using the fact that 
, we can write:
Clearly 
is a square, and as 
, and 
are primes, none of the other four 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. 
