Difference between revisions of "1998 AHSME Problems/Problem 22"
(→Solution 1) |
(→Solution 1) |
||
Line 12: | Line 12: | ||
<cmath>\log_{k} 100! = \frac{\log 100!}{\log k}</cmath> | <cmath>\log_{k} 100! = \frac{\log 100!}{\log k}</cmath> | ||
Thus (you might recognize this [[identity]] directly) | Thus (you might recognize this [[identity]] directly) | ||
− | < | + | <cmath>\frac{1}{\log_k 100!} = \frac{\log k}{\log 100!}</cmath> |
Thus the sum is | Thus the sum is | ||
<cmath>\left(\frac{1}{\log 100!}\right)(\log 1 + \log 2 + \cdots + \log 100) = \frac{1}{\log 100!} \cdot \log 100! = 1 \Rightarrow \mathrm{(C)}</cmath> | <cmath>\left(\frac{1}{\log 100!}\right)(\log 1 + \log 2 + \cdots + \log 100) = \frac{1}{\log 100!} \cdot \log 100! = 1 \Rightarrow \mathrm{(C)}</cmath> |
Revision as of 09:34, 22 April 2016
Contents
Problem
What is the value of the expression
Solution 1
By the change-of-base formula, Thus (you might recognize this identity directly) Thus the sum is
Solution 2
Since ,
We add:
See also
1998 AHSME (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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.