Difference between revisions of "2008 AIME II Problems/Problem 4"
I like pie (talk | contribs) m (→Solution) |
m |
||
| Line 9: | Line 9: | ||
In base <math>3</math>, we find that <math>\overline{2008}_{10} = \overline{2202101}_{3}</math>. In other words, | In base <math>3</math>, we find that <math>\overline{2008}_{10} = \overline{2202101}_{3}</math>. In other words, | ||
<center><math>2008 = 2 \cdot 3^{6} + 2 \cdot 3^{5} + 2 \cdot 3^3 + 1 \cdot 3^2 + 1 \cdot 3^0</math></center> | <center><math>2008 = 2 \cdot 3^{6} + 2 \cdot 3^{5} + 2 \cdot 3^3 + 1 \cdot 3^2 + 1 \cdot 3^0</math></center> | ||
| − | In order to rewrite as a sum of perfect powers of <math>3</math>, we can use that <math>2 \cdot 3^k = 3^{k+1} - 3^k</math>: | + | In order to rewrite as a sum of perfect powers of <math>3</math>, we can use the fact that <math>2 \cdot 3^k = 3^{k+1} - 3^k</math>: |
<center><math>2008 = (3^7 - 3^6) + (3^6-3^5) + (3^4 - 3^3) + 3^2 + 3^0 = 3^7 - 3^5 + 3^4 - 3^3 + 3^2 + 3^0</math></center> | <center><math>2008 = (3^7 - 3^6) + (3^6-3^5) + (3^4 - 3^3) + 3^2 + 3^0 = 3^7 - 3^5 + 3^4 - 3^3 + 3^2 + 3^0</math></center> | ||
The answer is <math>7+5+4+3+2+0 = \boxed{021}</math>. | The answer is <math>7+5+4+3+2+0 = \boxed{021}</math>. | ||
Revision as of 10:33, 8 June 2008
Problem
There exist 
unique nonnegative integers 
and unique integers 
(
) with each either 
or 
such that
![\[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\]](http://latex.artofproblemsolving.com/8/5/8/8583fcf8b5e36851cf3d1546038070e8b70e6061.png)
Find 
.
Solution
In base 
, we find that 
. In other words,
In order to rewrite as a sum of perfect powers of , we can use the fact that 
:
The answer is 
.
See also
| 2008 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
