1991 AIME Problems/Problem 1
From AoPSWiki
Problem
Find 
if 
and 
are positive integers such that
Contents |
Solution
Solution 1
Define 
and 
. Then 
and 
. Solving these two equations yields a quadratic: 
, which factors to 
. Either 
and 
or 
and 
. For the first case, it is easy to see that 
can be 
(or vice versa). In the second case, since all factors of 
must be 
, no two factors of can sum greater than 
, and so there are no integral solutions for . The solution is 
.
Solution 2
Since 
, this can be factored to 
. As 
and 
are integers, the possible sets for (ignoring cases where 
since it is symmetrical) are 
. The second equation factors to 
. The only set with a factor of 
is , and checking shows that it is our solution.
See also
| 1991 AIME (Problems • Resources) | ||
| Preceded by First question | Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
