Difference between revisions of "2002 AMC 12A Problems/Problem 14"
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First, note that <math>2002 = 11 \cdot 13 \cdot 14</math>. | First, note that <math>2002 = 11 \cdot 13 \cdot 14</math>. | ||
| − | Using the fact that for any base we have <math>\log a + \log b = \log ab</math>, we get that <math>N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{(D) 2}</math>. | + | Using the fact that for any base we have <math>\log a + \log b = \log ab</math>, we get that <math>N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{(D) N=2}</math>. |
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/YUdemyBf2yI | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2002|ab=A|num-b=13|num-a=15}} | {{AMC12 box|year=2002|ab=A|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:43, 25 August 2022
Contents
Problem
For all positive integers 
, let 
. Let 
. Which of the following relations is true?
Solution
First, note that 
.
Using the fact that for any base we have 
, we get that 
.
Video Solution
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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. 
