1992 IMO Problems/Problem 1
Problem
Find all integers , , satisfying such that is a divisor of .
Solution
With it implies that , ,
Therefore,
which for gives: , which gives :
for gives: , which gives :
for gives: , which gives :
Substituting those inequalities into the original inequality gives:
Since needs to be integer,
then or
Case 1:
Case 1, subcase :
gives: which has no solution because is even.
Case 1, subcase :
and provides solution
Case 2:
Case 2, subcase :
and provides solution
Case 2, subcase :
Since ) mod and mod , then there is no solution for this subcase.
Now we verify our two solutions:
when
and
Since is a factor of , this solutions is correct.
when
and
Since is a factor of , this solutions is also correct.
The solutions are: and
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1992 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |