1998 AHSME Problems/Problem 22

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:

\[\log_{100!} 1 +\log_{100!} 2 +\log_{100!} 3 +\cdots + \log_{100!} 100=\log_{100!}100!=1 \Rightarrow \mathrm{(C)}\]

See also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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