2016 AIME I Problems/Problem 11
Contents
Problem
Let be a nonzero polynomial such that for every real , and . Then , where and are relatively prime positive integers. Find .
Solution 1
We substitute into to get . Since we also have that , we have that and . We can also substitute , , and into to get that , , and . This leads us to the conclusion that and .
We next use finite differences to find that is a cubic polynomial. Thus, must be of the form of . It follows that ; we now have a system of equations to solve. We plug in , , and to get
We solve this system to get that , , and . Thus, . Plugging in , we see that . Thus, , , and our answer is .
Solution 2
So from the equation we see that divides and divides so we can conclude that and divide . This means that and are roots of . Plug in and we see that so is also a root.
Suppose we had another root that is not those . Notice that the equation above indicates that if is a root then and is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.
That means . We can use to get . Plugging in is now trivial and we see that it is so our answer is
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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