Difference between revisions of "2005 AIME II Problems/Problem 1"
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We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. | We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. | ||
| − | With a little work we realize the factorization <math>8 \cdot 9 \cdot 10</math>, so <math>n - 3 = 10</math> and <math>n = | + | With a little work we realize the factorization <math>8 \cdot 9 \cdot 10</math>, so <math>n - 3 = 10</math> and <math>n = \boxed{013}</math>. |
== See Also == | == See Also == | ||
Revision as of 17:42, 17 June 2011
Problem
A game uses a deck of 
different cards, where is an integer and 
The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find 
Solution
The number of ways to draw six cards from is given by the binomial coefficient 
.
The number of ways to choose three cards from is 
.
We are given that 
, so 
.
Cancelling like terms, we get 
.
We must find a factorization of the left-hand side of this equation into three consecutive integers.
With a little work we realize the factorization 
, so 
and 
.
See Also
| 2005 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
