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 , and the normal sum of all the numbers is , the positive numbers must add to and negative ones must add to . 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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