Difference between revisions of "2022 AIME II Problems/Problem 6"
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==Problem== | ==Problem== | ||
− | + | ur mom | |
==Solution 1== | ==Solution 1== |
Revision as of 13:41, 12 August 2023
Contents
Problem
ur mom
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
its biggest possible value.) That means
, and
.
and
, and subtracting them
.
.
~heheman
Video Solution by The Power of Logic
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.