Difference between revisions of "2002 AMC 12B Problems/Problem 22"
(→Solution 2) |
(→Solution 3) |
||
Line 25: | Line 25: | ||
== Solution 3 == | == Solution 3 == | ||
− | Note that <math>a_2 = \frac{1}{\log_2 | + | Note that <math>a_2 = \frac{1}{\log_2 2002}</math>. <math>1</math> is also equal to <math>\log_2 2</math>. So <math>a_2 = \frac{\log_2 2}{\log_2 2002}</math>. By the change of bases formula, <math>a_2 = \log_{2002} 2</math>. Following the same reasoning, <math>a_3 = \log_{2002} 3</math>, <math>a_4 = \log_{2002} 4</math> and so on. |
− | <cmath>b = \log_{ | + | <cmath>b = \log_{2002} 2 + \log_{2002} 3 + .....+ \log_{2002} 5 = \log_{2002} 5! = \log_{2002} 120</cmath> |
− | Now solving for <math>c</math>, we see that it equals <math>\log_{ | + | Now solving for <math>c</math>, we see that it equals <math>\log_{2002} (10\cdot 11 \cdot 12 \cdot 13 \cdot 14)</math> |
− | <cmath>b-c = \log_{ | + | <cmath>b-c = \log_{2002} 120 - \log_{2002} 240240 \rightarrow \log_{2002} \frac{1}{2002} = \boxed{-1}</cmath> |
~YBSuburbanTea | ~YBSuburbanTea |
Latest revision as of 18:26, 13 January 2022
Problem
For all integers greater than , define . Let and . Then equals
Solution
By the change of base formula, . Thus
Solution 2
Note that . Thus . Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following:
~yofro
Solution 3
Note that . is also equal to . So . By the change of bases formula, . Following the same reasoning, , and so on.
Now solving for , we see that it equals
~YBSuburbanTea
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.