2005 AIME II Problems/Problem 13
Problem
Let be a polynomial with integer coefficients that satisfies and Given that has two distinct integer solutions and find the product
Solution
We define , noting that it has roots at and . Hence . In particular, this means that . Therefore, satisfy , where , , and are integers. This cannot occur if or because the product will either be too large or not be a divisor of . We find that and are the only values that allow to be a factor of . Hence the answer is .
Solution 2
We know that so has two distinct solutions so is at least quadratic. Let us first try this problem out as if is a quadratic polynomial. Thus because where are all integers. Thus where are all integers. We know that or and or . By doing we obtain that or or . Thus . Now we know that , we have or which makes . Thus . By Vieta's formulas, we know that the sum of the roots() is equal to 41 and the product of the roots() is equal to . Because the roots are integers has to be an integer, so . Thus the product of the roots is equal to one of the following: . Testing every potential product of the roots, we find out that the only product that can have divisors that sum up to is .
-David Camacho
Solution 3
We have . Using the property that whenever and are distinct integers, we get and Since and , we must have We look for two divisors of that differ by ; we find that and satisfies these conditions. Therefore, either , giving , or , giving . From this, we conclude that and that .
~ Alcumus (Solution)
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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