Difference between revisions of "2016 AIME I Problems/Problem 11"
Mathgeek2006 (talk | contribs) |
Mathgeek2006 (talk | contribs) m (→Solution 4) |
||
Line 27: | Line 27: | ||
Applying this repeatedly, we find that | Applying this repeatedly, we find that | ||
<cmath>P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).</cmath> | <cmath>P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).</cmath> | ||
− | Therefore, as <math>P(2)=4</math>, we find <math>P(n+1)=\frac{2}{3}(n+2)(n+1)n</math> for all positive integers <math>n\ge2</math>. This cubic polynomial matches the values <math>P(n+1)</math> for infinitely many numbers, hence the two | + | Therefore, as <math>P(2)=4</math>, we find <math>P(n+1)=\frac{2}{3}(n+2)(n+1)n</math> for all positive integers <math>n\ge2</math>. This cubic polynomial matches the values <math>P(n+1)</math> for infinitely many numbers, hence the two polynomials are identically equal. In particular, <math>P\left(\frac72\right)=\frac23\cdot\frac92\cdot\frac72\cdot\frac52=\frac{105}{4}</math>, and the answer is <math>\boxed{109}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=10|num-a=12}} | {{AIME box|year=2016|n=I|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:29, 4 March 2016
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
Solution 3
Although this may not be the most mathematically rigorous answer, we see that . Using a bit of logic, we can make a guess that has a factor of , telling us has a factor of . Similarly, we guess that has a factor of , which means has a factor of . Now, since and have so many factors that are off by one, we may surmise that when you plug into , the factors "shift over," i.e. , which goes to . This is useful because these, when divided, result in . If , then we get and , . This gives us and , and at this point we realize that there has to be some constant multiplied in front of the factors, which won't affect our fraction but will give us the correct values of and . Thus , and we utilize to find . Evaluating is then easy, and we see it equals , so the answer is
Solution 4
As above, we find that . Now for integers , we know that Applying this repeatedly, we find that Therefore, as , we find for all positive integers . This cubic polynomial matches the values for infinitely many numbers, hence the two polynomials are identically equal. In particular, , and the answer is .
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.