1998 AIME Problems/Problem 2
Problem
Find the number of ordered pairs of positive integers that satisfy and .
Contents
Solution
Solution 1
Pick's theorem states that:
The conditions give us four inequalities: , , , . These create a quadrilateral, whose area is of the 30 by 30 square it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of the area of the 30 by 30 square.
So . we can calculate by just counting. Ignoring the vertices, the top and right sides have 14 lattice points, and the two diagonals each have 14 lattice points (for the top diagonal, every value of corresponds with an integer value of as and vice versa for the bottom, so and there are 14 values for x not counting vertices). Adding the four vertices, there are 60 points on the borders.
Since the inequalities also include the equals case, we include the boundaries, which gives us ordered pairs. However, the question asks us for positive integers, so doesn't count; hence, the answer is .
Solution 2
First, note that all pairs of the form , work.
Now, considered the ordered pairs with , so that is automatically satisfied. Since , there are possible values of . Hence, given , there are values of possible for which and the above conditions are satisfied. But , so this only works for . Thus, there are
ordered pairs. For , must follow . Hence, there are possibilities for , and there are
ordered pairs.
By symmetry, there are also ordered pairs with and the above criteria satisfied.
Hence, the total is
Solution 3
Multiplying both sides by 2 yields:
Then the two inequalities can be merged to form the following inequality:
Additionally, we must ensure that
Therefore we must find pairs that satisfy the inequality above. A bit of trial and error and observing patterns leads to the answer .
It should be noted that the cases for and should be considered separately in order to ensure that .
Solution 4 - Unrigorous engineers induction solution
We will try out small cases.
By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the difference between adjacent terms (1,3,3,5,5,...). We suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480.
-Alexlikemath
Solution 5 - Counting Head on
Notice and both must be equal or less than 30. The inequalities given have no complicated qualities. We can recompile them by understanding: Two times the larger integer will also be larger than the smaller integer; Two times the smaller integer is greater or equal to the greater integer if and only if the greater integer is less or equal to the double of the smaller integer. Knowing this, we create a chart. We will first solve without order, then multiply pairs by 2 at the end. can be 1-30, so we'll start with 1. The only possible value for is 1. For , can be 2-4. For , can be 3-6. There is an obvious pattern here. For every integer after 1, the possible values for will be numbers -. This predictably ends at because will reach 30. When , then the number of possible values of will begin to drop again, equaling the amount when . Then when we finally sum the group together, for to there are 104 pairs with 2 distinct values, and 13 values with congruent values. These will not be multiplied by 2 later on. to gives the same amount. Then and gives 16 and 2 values respectively, with two congruent values each. Finally, and give 3 and 1 respectively.
Sum them together and you will get 480.
-jackshi2006
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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