Difference between revisions of "2000 AIME II Problems/Problem 1"
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Therefore our answer is <math>1 + 6 = \boxed{007}</math>. | Therefore our answer is <math>1 + 6 = \boxed{007}</math>. | ||
| + | == Solution 3 == | ||
| + | We know that <math>2 = \log_4{16}</math> and <math>3 = \log_5{125}</math>, and by base of change formula, <math>\log_a{b} = \frac{\log_c{b}}{\log_c{a}}</math>. Lastly, notice <math>\log a + \log b = \log ab</math> for all bases. | ||
| + | <cmath> | ||
| + | \begin{align*} | ||
| + | \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} = \log_{2000^6}{16} + \log_{2000^6}{125} = \log_{2000^6}{2000} = \frac16 \implies \boxed{007} \end{align*}</cmath> | ||
| + | |||
| + | <math>\bold{Solution}</math> <math>\bold{written}</math> <math>\bold{by}</math> | ||
| + | |||
| + | ~ <math>\bold{PaperMath}</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 16:27, 16 August 2023
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 
.
Solution 3
We know that 
and 
, and by base of change formula, 
. Lastly, notice 
for all bases.
~ 
| 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. 
