Difference between revisions of "2003 AIME I Problems/Problem 1"
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where <math> k </math> and <math> n </math> are [[positive integer]]s and <math> n </math> is as large as possible, find <math> k + n. </math> | where <math> k </math> and <math> n </math> are [[positive integer]]s and <math> n </math> is as large as possible, find <math> k + n. </math> | ||
− | == Solution == | + | ==Solution == |
− | + | Note that<cmath>{{\left((3!)!\right)!}\over{3!}}= | |
+ | {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.</cmath>Because <math>120\cdot719!<720!</math>, we can conclude that <math>n < 720</math>. Thus, the maximum value of <math>n</math> is <math>719</math>. The requested value of <math>k+n</math> is therefore <math>120+719=\boxed{839}</math>. | ||
− | + | ~yofro | |
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== See also == | == See also == | ||
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 06:27, 21 November 2023
Problem
Given that
where and are positive integers and is as large as possible, find
Solution
Note thatBecause , we can conclude that . Thus, the maximum value of is . The requested value of is therefore .
~yofro
See also
2003 AIME I (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 |
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