1963 AHSME Problems/Problem 31
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 the problem as following,
We can apply hit and trial for the first solution 
= 
and 
=. then the general solution of the given diophanitine equation will be = +
and = - 
. We know that we need only positive integer solutions. so + > 
and - > to get 
> 0 (applying Greatest integer function) also we can clearly see that $t_(min)_$ (Error compiling LaTeX. Unknown error_msg) = 0 so,t < 
(
/
). That implies ranges from to . Hence,the correct answer is , .
Solution by Geometry-Wizard
See Also
| 1963 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
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