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1998 AHSME Problems/Problem 22

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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!}?

\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 \...

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 (Problems)
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
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