1991 AIME Problems/Problem 1
Problem
Find if and are positive integers such that
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 correct.
Solution 3
Let , then we get the equations After finding the prime factorization of , it's easy to obtain the solution . Thus Note that if , the answer would exceed which is invalid for an AIME answer. ~ Nafer
Solution 4
From the first equation, we know . We factor the second equation as . Let and rearranging we get . We have two cases: (1) and OR (2) and . We find the former is true for . .
Solution 5
First, notice that you can factor as . From this, we notice that and is a common occurrence, so that lends itself to a simple solution by substitution. Let and . From this substitution, we get the following system: Solving that system gives us the following two pairs : and . The second one is obviously too big as is clearly out of bounds for an AIME problem (More on that later). Therefore, we proceed with substituting back the pair . This means that and Then, instead of solving the system, we can do a clever manipulation by squaring . Doing so, we get: We see that in this form, we can substitute everything in except for , which is the desired answer. Substituting, we get: so . (If we were to go with the pair , then the would be absurdly out of bounds)
~EricShi1685
See also
1991 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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