Difference between revisions of "2022 AIME II Problems/Problem 6"
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| + | ==Problem== | ||
| + | Let <math>x_1\leq x_2\leq \cdots\leq x_{100}</math> be real numbers such that <math>|x_1| + |x_2| + \cdots + |x_{100}| = 1</math> and <math>x_1 + x_2 + \cdots + x_{100} = 0</math>. Among all such <math>100</math>-tuples of numbers, the greatest value that <math>x_{76} - x_{16}</math> can achieve is <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | ==See Also== | ||
| + | {{AIME box|year=2022|n=II|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 22:14, 17 February 2022
Problem
Let 
be real numbers such that 
and 
. Among all such 
-tuples of numbers, the greatest value that 
can achieve is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
See Also
| 2022 AIME II (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 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
