2022 AIME II Problems/Problem 6
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 1, and the normal sum of all the numbers is 0, The positive numbers must add to 1/2 and negative ones must add to -1/2. To maximize x76 - x16, we must make x76 as big as possible and x16 as small as possible. We can do this by making x1 + x2 + x3 ... x16 = -1/2, where x1 = x2 = x3 = ... = x16 (because that makes x16 the smallest possible value), and x76 + x77 + x78 + ... + x100 = 1/2, where similarly x76 = x77 = ... = x100 (because it makes x76 its biggest possible value.) That means 16(x16) = -1/2, and 25(x76) = 1/2. x16 = -1/32 and x76 = 1/50, 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 |
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