Difference between revisions of "2016 AIME II Problems/Problem 6"
m |
m |
||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
For polynomial <math>P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}</math>, define | For polynomial <math>P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}</math>, define | ||
<math>Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}</math>. | <math>Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}</math>. | ||
Revision as of 16:18, 22 March 2018
Contents
Problem
For polynomial 
, define
.
Then 
, where 
and 
are relatively prime positive integers. Find 
.
Solution 1
Note that all the coefficients of odd-powered terms is an odd number of odd degree terms multiplied together, and all coefficients of even-powered terms have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to 
, so the desired answer is 
.
Solution by Shaddoll
Solution 2
We are looking for the sum of the absolute values of the coefficients of 
. By defining 
, and defining 
, we have made it so that all coefficients in 
are just the positive/absolute values of the coefficients of . .
To find the sum of the absolute values of the coefficients of , we can just take the sum of the coefficients of . This sum is equal to
![\[Q'(1) = P'(1)P'(1)P'(1)P'(1)P'(1) = \left(1+\frac{1}{3}+\frac{1}{6}\right)^5 = \frac{243}{32},\]](http://latex.artofproblemsolving.com/5/0/7/507271d1f2043b3d858e5bd017b7e309a552eb79.png)
so our answer is 
.
Solution by ExuberantGPACN
See also
| 2016 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
