Difference between revisions of "2022 AIME II Problems/Problem 6"
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==Solution 1== | ==Solution 1== | ||
− | To find the greatest value of <math>x_{76} - x_{16}</math>, <math>x_{76}</math> must be as large as possible, and <math>x_{16}</math> must be as small as possible. If <math>x_{76}</math> is as large as possible, <math> | + | To find the greatest value of <math>x_{76} - x_{16}</math>, <math>x_{76}</math> must be as large as possible, and <math>x_{16}</math> must be as small as possible. If <math>x_{76}</math> is as large as possible, <math>x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0</math>. If <math>x_{16}</math> is as small as possible, <math>x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0</math>. The other numbers between <math>x_{16}</math> and <math>x_{76}</math> equal to <math>0</math>. Let <math>a = x_{76}</math>, <math>b = x_{76}</math>. Substituting <math>a</math> and <math>b</math> into <math>|x_1| + |x_2| + \cdots + |x_{100}| = 1</math> and <math>x_1 + x_2 + \cdots + x_{100} = 0</math> we get: |
<cmath>25a - 16b = 1</cmath> | <cmath>25a - 16b = 1</cmath> | ||
<cmath>25a + 16b = 0</cmath> | <cmath>25a + 16b = 0</cmath> |
Revision as of 06:02, 19 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 1
To find the greatest value of ,
must be as large as possible, and
must be as small as possible. If
is as large as possible,
. If
is as small as possible,
. The other numbers between
and
equal to
. Let
,
. Substituting
and
into
and
we get:
,
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.