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 = 380 and
=1. then the general solution of the given diophanitine equation will be x =
+3t and y =
- 2t. we know that we need only positive integer solutions. so 380 + 3t > 0 and 1-2t >0 to get t>0 (applying Greatest integer function) also we can clearly see that
= 0 so,t < GIF(383/3). That implies t ranges from 0 to 127. Hence,the correct answer is 127,
..
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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