2022 AIME II Problems/Problem 6
Contents
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 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 | |
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