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

Latest revision as of 20:47, 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 $1+3+3^2 + \cdots + 3^{n} = \displaystyle\frac{3^{n+1}-1}{2}.$

@@ 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>
 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 ==

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

Latest revision as of 20:47, 21 September 2020

Problem

Solution

See also