Difference between revisions of "1998 AIME Problems/Problem 2"
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It should be noted that the cases for <math>x\le15</math> and <math>x>15</math> should be considered separately in order to ensure that <math>2y < 60</math>. | It should be noted that the cases for <math>x\le15</math> and <math>x>15</math> should be considered separately in order to ensure that <math>2y < 60</math>. | ||
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+ | ===Solution 4 - Unrigorous engineers induction solution=== | ||
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+ | We will try out small cases. | ||
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+ | 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. | ||
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+ | -Alexlikemath | ||
== See also == | == See also == |
Latest revision as of 00:30, 3 June 2020
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
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 | ||
All AIME Problems and Solutions |
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