Difference between revisions of "Mock AIME 2 Pre 2005 Problems/Problem 9"
(→Solution) |
(→Solution) |
||
| Line 14: | Line 14: | ||
=665280 | =665280 | ||
=>\boxed{280}</cmath> | =>\boxed{280}</cmath> | ||
| + | |||
| + | -MRGORILLA | ||
Revision as of 15:01, 31 December 2020
Solution
We begin by determining the value of 
. Experimenting, we find the first few 
s:
![\[k_{1} = 3, k_{2} = {3^2}, k_{3} = {3^2}+3, k_{4} = {3^3}...\]](http://latex.artofproblemsolving.com/e/c/2/ec25b2f4e8bcb96d11cc80eca12a5141e930f954.png)
We observe that because 
, will be determined by the base 2 expansion of i. Specifically, every 1 in the 
s digit of the expansion corresponds to adding 
to . Since 
base 2,
![\[k_{1997}={3^{11}}+{3^{10}}+{3^9}+{3^8}+{3^7}+{3^4}+{3^3}+{3^1}.\]](http://latex.artofproblemsolving.com/e/f/2/ef2776c115da46de44ccbfb1bfced24494f29fab.png)
Now we look for ways to attain an element with degree . Since each sum of powers of 3 is unique, there is only one; namely, take the x element for every binomial with a degree of one of the added powers of 3 in , and the 1 for all else. Finally, since the coefficients of the x elements are equal to the degree to which the 3 is raised, we conclude
![\[a_{1997}=11*10*9*8*7*4*3*1 =665280 =>\boxed{280}\]](http://latex.artofproblemsolving.com/9/0/1/90135463d4b1fd64622501b659e0e2439e635a59.png)
-MRGORILLA
