Difference between revisions of "2005 AIME II Problems/Problem 1"
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A game uses a deck of <math> n </math> different cards, where <math> n </math> is an integer and <math> n \geq 6. </math> 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 <math> n. </math> | A game uses a deck of <math> n </math> different cards, where <math> n </math> is an integer and <math> n \geq 6. </math> 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 <math> n. </math> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/IRyWOZQMTV8?t=150 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== Solution == | == Solution == | ||
Line 13: | Line 18: | ||
Cancelling like terms, we get <math>(n - 3)(n - 4)(n - 5) = 720</math>. | Cancelling like terms, we get <math>(n - 3)(n - 4)(n - 5) = 720</math>. | ||
− | We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. Since 720 is close to <math>9^3=729</math>, we try 8, 9, and 10, which works, so <math>n - 3 = 10</math> and <math>n = \boxed{ | + | We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. Since 720 is close to <math>9^3=729</math>, we try 8, 9, and 10, which works, so <math>n - 3 = 10</math> and <math>n = \boxed{13}</math>. |
== See Also == | == See Also == | ||
{{AIME box|year=2005|n=II|before=First Question|num-a=2}} | {{AIME box|year=2005|n=II|before=First Question|num-a=2}} | ||
− | [[Category:Introductory Combinatorics Problems]] | + | [[Category:Introductory Combinatorics Problems, Introductory Algebra Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:54, 31 July 2023
Contents
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
Video Solution
https://youtu.be/IRyWOZQMTV8?t=150
~ pi_is_3.14
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. Since 720 is close to , we try 8, 9, and 10, which works, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.