1963 AHSME Problems/Problem 31
Contents
[hide]Problem
The number of solutions in positive integers of is:
Solution 1
Solving for in the equation yields . Solving the inequality results in . From the two conditions, can be an odd number from to , so there are solutions where and are integers. The answer is .
Solution 2
We will prove that is an odd number by contradiction. If is even, then we know that where is some integer. However, this immediately assumes that which is impossible. therefore must ben odd. then we can easily prove that .....
Solution 3
We can solve for first solution by applying extended euclids division algorithm or we can apply hit and trial for the first solution to get = and = . then the general solution of the given diophanitine equation will be = + and = - . Since we need only positive integer solutions So we solve + and - to get (applying Greatest integer function) also we can clearly see that so,t (/). That implies ranges from to . Hence,the correct answer is , .
~Geometry-Wizard
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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