Difference between revisions of "2008 AIME II Problems/Problem 4"

(Solution)
 
Line 12: Line 12:
 
<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>.
 +
 +
Note : Solution by bounding is also possible, namely using the fact that <math>1+3+3^2 + \cdots + 3^{n} = \displaystyle\frac{3^{n+1}-1}{2}.</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 21:47, 21 September 2020

Problem

There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ ($1\le k\le r$) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$.

Solution

In base $3$, we find that $\overline{2008}_{10} = \overline{2202101}_{3}$. In other words,

$2008 = 2 \cdot 3^{6} + 2 \cdot 3^{5} + 2 \cdot 3^3 + 1 \cdot 3^2 + 1 \cdot 3^0$

In order to rewrite as a sum of perfect powers of $3$, we can use the fact that $2 \cdot 3^k = 3^{k+1} - 3^k$:

$2008 = (3^7 - 3^6) + (3^6-3^5) + (3^4 - 3^3) + 3^2 + 3^0 = 3^7 - 3^5 + 3^4 - 3^3 + 3^2 + 3^0$

The answer is $7+5+4+3+2+0 = \boxed{021}$.

Note : Solution by bounding is also possible, namely using the fact that $1+3+3^2 + \cdots + 3^{n} = \displaystyle\frac{3^{n+1}-1}{2}.$

See also

2008 AIME II (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS