Difference between revisions of "2000 AIME II Problems/Problem 1"
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Alternatively, we could've noted that, because <math>\frac 1{\log_a{b}} = \log_b{a}</math> | Alternatively, we could've noted that, because <math>\frac 1{\log_a{b}} = \log_b{a}</math> | ||
| − | <cmath>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} = | + | <cmath> |
| + | \begin{align*} | ||
| + | \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} &= 2 \cdot \frac{1}{\log_4{2000^6}} + 3\cdot \frac {1}{\log_5{2000^6} }\\ | ||
| + | &=2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} \\ | ||
| + | &={\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}}\\ | ||
| + | &={\log_{2000^6}{4^2 \cdot 5^3}}\\ | ||
| + | &={\log_{2000^6}{2000}}\\ | ||
| + | &= {\frac{1}{6}}.\end{align*}</cmath> | ||
| − | + | Therefore our answer is <math>1 + 6 = \boxed{7}</math>. | |
| − | |||
| − | |||
| − | |||
| − | Therefore our answer is <math>1 + 6 = \boxed{7}</math> | ||
{{AIME box|year=2000|n=II|before=First Question|num-a=2}} | {{AIME box|year=2000|n=II|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:41, 13 March 2015
Contents
Problem
The number
can be written as 
where 
and 
are relatively prime positive integers. Find 
.
Solution
Solution 1
Therefore, 
Solution 2
Alternatively, we could've noted that, because 
Therefore our answer is 
.
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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. 
