2015 AIME II Problems/Problem 8
Problem
Let and be positive integers satisfying . The maximum possible value of is , where and are relatively prime positive integers. Find .
Solution 1
Let us call the quantity as for convenience. Knowing that and are positive integers, we can legitimately rearrange the given inequality so that is by itself, which makes it easier to determine the pairs of that work. Doing so, we have Now, observe that if we have that , regardless of the value of . If , we have the same result: that , regardless of the value of . Hence, we want to find pairs of positive integers existing such that neither nor is equal to , and that the conditions given in the problem are satisfied in order to check that the maximum value for is not .
To avoid the possibility that , we want to find values of such that . If we do this, we will have that , where is greater than , and this allows us to choose values of greater than . Again, since is a positive integer, and we want , we can legitimately multiply both sides of by to get . For , we have that , so the only possibility for greater than is obviously . Plugging these values into , we have that . For , we have that . Plugging and in for yields the same result of , but plugging and into yields that . Clearly, is the largest value we can have for , so our answer is .
(Technically, we would have to find that b > 1 before dividing both sides of the inequality by 2b - 3, but otherwise this solution is correct)
Solution 2 (best solution)
and each cannot be even or else and will not be integers
(; ).
Solution 3
Notice that for to be maximized, has to be maximized. We simplify as above to , which is . To maximize, has to be as close to as possible, making close to . Because and are positive integers, , and checking back gives as the maximum or the other way around, which the answer is thus .
Solution 4
Guess and check a few values of and you will find two things. One, that the highest values of and are the closest to . Two, that the pair is the highest possible value of . So plugging in and we get = = , and . Solution by Helloitsaaryan
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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