Difference between revisions of "2008 AIME II Problems/Problem 4"
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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>. | ||
+ | |||
+ | 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 == | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:47, 21 September 2020
Problem
There exist unique nonnegative integers and unique integers () with each either or such that 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 .
Note : Solution by bounding is also possible, namely using the fact that
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.