Difference between revisions of "2022 AIME II Problems/Problem 6"
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− | Because the absolute value sum of all the numbers is <math>1</math>, and the normal sum of all the numbers is <math>0</math>, the positive numbers must add to <math>\frac12</math> and negative ones must add to <math>-\ | + | Because the absolute value sum of all the numbers is <math>1</math>, and the normal sum of all the numbers is <math>0</math>, the positive numbers must add to <math>\frac12</math> and negative ones must add to <math>-\dfrac12</math>. To maximize <math>x_{76} - x_{16}</math>, we must make <math>x_{76}</math> as big as possible and <math>x_{16}</math> as small as possible. We can do this by making <math>x_1 + x_2 + x_3 \dots x_{16} = -\dfrac{1}{2}</math>, where <math>x_1 = x_2 = x_3 = \dots = x_{16}</math> (because that makes <math>x_{16}</math> the smallest possible value), and <math>x_{76} + x_{77} + x_{78} + \dots + x_{100} = \dfrac{1}{2}</math>, where similarly <math>x_{76} = x_{77} = \dots = x_{100}</math> (because it makes x_{76} its biggest possible value.) That means <math>16(x_{16}) = -\dfrac{1}{2}</math>, and <math>25(x_{76}) = \dfrac{1}{2}</math>. <math>x_{16} = -\dfrac{1}{32}</math> and <math>x_{76} = \dfrac{1}{50}</math>, and subtracting them 1/50 - (-1/32) = 41/400. 41 + 400 = 841. |
~heheman | ~heheman |
Revision as of 19:35, 12 May 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:
,
.
Solution 2
Define to be the sum of all the negatives, and
to be the sum of all the positives.
Since the sum of the absolute values of all the numbers is ,
.
Since the sum of all the numbers is ,
.
Therefore, , so
and
since
is negative and
is positive.
To maximize , we need to make
as small of a negative as possible, and
as large of a positive as possible.
Note that is greater than or equal to
because the numbers are in increasing order.
Similarly, is less than or equal to
.
So we now know that is the best we can do for
, and
is the least we can do for
.
Finally, the maximum value of , so the answer is
.
(Indeed, we can easily show that ,
, and
works.)
~inventivedant
Solution 3
Because the absolute value sum of all the numbers is , and the normal sum of all the numbers is
, the positive numbers must add to
and negative ones must add to
. To maximize
, we must make
as big as possible and
as small as possible. We can do this by making
, where
(because that makes
the smallest possible value), and
, where similarly
(because it makes x_{76} its biggest possible value.) That means
, and
.
and
, and subtracting them 1/50 - (-1/32) = 41/400. 41 + 400 = 841.
~heheman
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.