2000 AIME I Problems/Problem 6
Because , we only consider .
For simplicity, we can count how many valid pairs of that satisfy our equation.
The maximum that can be is because must be an integer (this is because , an integer). Then , and we continue this downward until , in which case . The number of pairs of , and so is then .
Let = and = , where and are positive.
This makes counting a lot easier since now we just have to find all pairs that differ by 2.
Because , then we can use all positive integers less than 1000 for and .
We know that because , we get .
We can count even and odd pairs separately to make things easier*:
This makes 499 odd pairs and 498 even pairs, for a total of pairs.
Note: We are counting the pairs for the values of and , which, when squared, translate to the pairs of we are trying to find.
Since the arithmetic mean is 2 more than the geometric mean, . We can multiply by 2 to get . Subtracting 4 and squaring gives
Notice that , so the problem asks for solutions of Since the left hand side is a perfect square, and 16 is a perfect square, must also be a perfect square. Since , must be from to , giving at most 999 options for .
However if , you get , which has solutions and . Both of those solutions are not less than , so cannot be equal to 1. If , you get , which has 2 solutions, , and . 16 is not less than 4, and cannot be 0, so cannot be 4. However, for all other , you get exactly 1 solution for , and that gives a total of pairs.
Solution 4 (Similar to Solution 3)
Rearranging our conditions to
Now, let Plugging this back into our expression, we get
There, a unique value of is formed for every value of . However, we must have
Therefore, there are only pairs of
Solution by Williamgolly
First we see that our condition is . Then we can see that . From trying a simple example to figure out conditions for , we want to find so we can isolate for . From doing the example we can note that we can square both sides and subtract : (note it is negative because . Clearly the square root must be an integer, so now let . Thus . Thus . We can then find , and use the quadratic formula on to ensure they are and respectively. Thus we get that can go up to 999 and can go down to , leaving possibilities for .
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