Difference between revisions of "1998 AHSME Problems/Problem 22"
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<cmath>\frac{1}{\log_k 100!} = \frac{\log k}{\log 100!}</cmath> | <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)( | + | <cmath>\left(\frac{1}{\log 100!}\right)(\log 2 + \cdots + \log 100) = \frac{1}{\log 100!} \cdot \log 100! = 1 \Rightarrow \mathrm{(C)}</cmath> |
=== Solution 2 === | === Solution 2 === | ||
Revision as of 09:40, 22 April 2016
Contents
Problem
What is the value of the expression
![\[\frac{1}{\log_2 100!} + \frac{1}{\log_3 100!} + \frac{1}{\log_4 100!} + \cdots + \frac{1}{\log_{100} 100!}?\]](http://latex.artofproblemsolving.com/e/6/3/e63e3b9ec89bd66181c2082ca2c3f5d3837f4853.png)
Solution 1
By the change-of-base formula,
![\[\log_{k} 100! = \frac{\log 100!}{\log k}\]](http://latex.artofproblemsolving.com/a/7/d/a7d308108bdf05a602b6a13f026286818a8ea2e5.png)
Thus (because if 
, then 
)
![\[\frac{1}{\log_k 100!} = \frac{\log k}{\log 100!}\]](http://latex.artofproblemsolving.com/1/8/0/1800fe46a3d13afe1c569797b2b1d3cc3889b266.png)
Thus the sum is
![\[\left(\frac{1}{\log 100!}\right)(\log 2 + \cdots + \log 100) = \frac{1}{\log 100!} \cdot \log 100! = 1 \Rightarrow \mathrm{(C)}\]](http://latex.artofproblemsolving.com/d/3/e/d3e8ba46c8d05286994117b6e4dd55f7daa8bb6f.png)
Solution 2
Since 
,
![\[\frac{1}{\log_{k}100!}=\frac{\log_{k}k}{\log_{k}100!}=\log_{100!} k\]](http://latex.artofproblemsolving.com/4/6/c/46cf706afa1d9d44629a51dc59bbc9945ee6840d.png)
We add:
![\[\log_{100!} 1 +\log_{100!} 2 +\log_{100!} 3 +\cdots + \log_{100!} 100=\log_{100!}100!=1 \Rightarrow \mathrm{(C)}\]](http://latex.artofproblemsolving.com/5/c/f/5cfa5d82f6aedc41a17f9f60149ce86adcf21b83.png)
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. 
