Difference between revisions of "1998 AHSME Problems/Problem 22"

Revision as of 15:35, 9 February 2008

Problem

What is the value of the expression $\frac{1}{\log_2 100!} + \frac{1}{\log_3 100!} + \frac{1}{\log_4 100!} + \cdots + \frac{1}{\log_{100} 100!}?$

$\mathrm{(A)}\ 0.01 \qquad\mathrm{(B)}\ 0.1 \qquad\mathrm{(C)}\ 1 \qquad\mathrm{(D)}\ 2 \qquad\mathrm{(E)}\ 10$

Solution

Solution 1

By the change-of-base formula, $\log_{k} 100! = \frac{\log 100!}{\log k}$ Thus (you might recognize this identity directly) $\frac{1}{\log_k 100!} = \frac{\log k}{\log 100!}$ Thus the sum is $\left(\frac{1}{\log 100!}\right)(\log 1 + \log 2 + \cdots + \log 100) = \frac{1}{\log 100!} \cdot \log 100! = 1 \Rightarrow \mathrm{(C)}$

Solution 2

Since $1=\log_{k} k$ ,

$\frac{1}{\log_{k}100!}=\frac{\log_{k}k}{\log_{k}100!}=\log_{100!} k$

We add:

$\log_{100!} 1 +\log_{100!} 1 +\log_{100!} 1 +\cdots + \log_{100!} 100=\log_{100!}100!=1 \Rightarrow \mathrm{(C)}$

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

@@ Line 20: / Line 20: @@
 Since <math>1=\log_{k} k</math>,
-<math></math>\frac{1}{\log_{k}100!}=\frac{\log_{k}k}{\log_{k}100!}=\log_{100!} k<math>
+<cmath>\frac{1}{\log_{k}100!}=\frac{\log_{k}k}{\log_{k}100!}=\log_{100!} k</cmath>
 We add:
-</math>\log_{100!} 1 +\log_{100!} 1 +\log_{100!} 1 +\cdots + \log_{100!} 100=\log_{100!}100!=1 \Rightarrow \mathrm{(C)}$
+<math>\log_{100!} 1 +\log_{100!} 1 +\log_{100!} 1 +\cdots + \log_{100!} 100=\log_{100!}100!=1 \Rightarrow \mathrm{(C)}</math>
 == See also ==

Page

Toolbox

Search