1998 AMC 12 Problem 22

Attempt Problem 22 of the 1998 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 12 solutions, or check the answer key.

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22.

What is the value of the expression 1log⁡2100!+1log⁡3100!+1log⁡4100!+⋯+1log⁡100100!? \begin{aligned} &\frac1{\log_2 100!}+\frac1{\log_3 100!}\\ &\quad+\frac1{\log_4 100!}+\cdots\\ &\qquad+\frac1{\log_{100}100!}? \end{aligned}
0.010.01
0.10.1
11
22
1010
Answer: C
Concepts:logarithmssums
Small hint:
Use the reciprocal identity 1log⁡bx=log⁡xb\frac{1}{\log_bx}=\log_xb
Big hint:
Combine the resulting logarithms of 2,3,…,1002,3,\ldots,100
Solution:
Each term equals log⁡100!b.\log_{100!}b. Therefore the sum is log⁡100!(2⋅3⋯100)=log⁡100!(100!)=1. \begin{aligned} &\log_{100!}(2\cdot3\cdots100)\\ &\qquad=\log_{100!}(100!)=1. \end{aligned} Thus C is correct.
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