Maths Olympiad Prep

Library / /316 of 520

Algebra Difficulty 3.3 AMC 10/12 Find the answer

For all integers nn greater than 11, define an=1log⁡n2002a_n = \frac{1}{\log_n 2002}. Let b=a2+a3+a4+a5b = a_2 + a_3 + a_4 + a_5 and c=a10+a11+a12+a13+a14c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}. Then b−cb- c equals

Pick one

Solution

By the change of base formula, an=1log⁡2002log⁡n=(1log⁡2002)log⁡na_n = \frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n. Thus
b−c=(1log⁡2002)(log⁡2+log⁡3+log⁡4+log⁡5−log⁡10−log⁡11−log⁡12−log⁡13−log⁡14)=(1log⁡2002)(log⁡2⋅3⋅4⋅510⋅11⋅12⋅13⋅14)=(1log⁡2002)log⁡2002−1=−(log⁡2002log⁡2002)=−1⇒(B)\begin{align*}b- c &= \left(\frac{1}{\log 2002}\right)(\log 2 + \log 3 + \log 4 + \log 5 - \log 10 - \log 11 - \log 12 - \log 13 - \log 14)\\ &= \left(\frac{1}{\log 2002}\right)\left(\log \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}\right)\\ &= \left(\frac{1}{\log 2002}\right) \log 2002^{-1} = -\left(\frac{\log 2002}{\log 2002}\right) = -1 \Rightarrow \mathrm{(B)}\end{align*}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.