Tallying Tiny Logs

Algebra Level 2

S = 2 × 2016 log 2 ( 2016 ! 2 ) + 3 × 2016 log 3 ( 2016 ! 3 ) + 4 × 2016 log 4 ( 2016 ! 4 ) + + 2016 × 2016 log 2016 ( 2016 ! 2016 ) S=\dfrac{2\times2016}{\log_{2}(2016!^{2})}+\dfrac{3\times2016}{\log_{3}(2016!^{3})}+\dfrac{4\times2016}{\log_{4}(2016!^{4})}+\dots+\dfrac{2016\times2016}{\log_{2016}(2016!^{2016})}

Find S S .

Notation : ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 2016.

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3 solutions

Chew-Seong Cheong
Jun 26, 2016

S = n = 2 2016 2016 n log n ( 2016 ! ) n = n = 2 2016 2016 n n log ( 2016 ! ) log n = n = 2 2016 2016 log n log ( 2016 ! ) = 2016 log ( 2016 ! ) log ( 2016 ! ) = 2016 \begin{aligned} S & = \sum_{n=2}^{2016} \frac {2016n}{\log_n (2016!)^n} = \sum_{n=2}^{2016} \frac {2016n}{\frac {n \log (2016!)}{\log n}} = \sum_{n=2}^{2016} \frac {2016 \log n}{\log (2016!)} =\frac {2016 \log (2016!)}{\log (2016!)} = \boxed{2016} \end{aligned}

Thanks you very much! It takes less steps to find the answer.By the way, you mean that :

S = n = 2 2016 2016 n log n ( 2016 ! ) n = n = 2 2016 2016 n n log ( 2016 ! ) log n = 2016 n = 2 2016 log n log ( 2016 ! ) = 2016 log ( 2016 ! ) log ( 2016 ! ) = 2016 \begin{aligned} S & = \sum_{n=2}^{2016} \frac {2016n}{\log_n (2016!)^n} = \sum_{n=2}^{2016} \frac {2016n}{\frac {n \log (2016!)}{\log n}} = \color{#3D99F6}{2016}\sum_{n=2}^{2016} \frac {\log n}{\log (2016!)} =\color{#3D99F6}{2016}\frac {\log (2016!)}{\log (2016!)} = \boxed{\color{#3D99F6}{2016}} \end{aligned}

Tommy Li - 4 years, 11 months ago

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Thanks. I was still thinking about the answer of another problem.

Chew-Seong Cheong - 4 years, 11 months ago
Tommy Li
Jun 25, 2016

S = 2 × 2016 log 2 ( 2016 ! ) 2 + 3 × 2016 log 3 ( 2016 ! ) 3 + 4 × 2016 log 4 ( 2016 ! ) 4 + + 2016 × 2016 log 2016 ( 2016 ! ) 2016 S=\dfrac{2\times2016}{\log_{2}(2016!)^{2}}+\dfrac{3\times2016}{\log_{3}(2016!)^{3}}+\dfrac{4\times2016}{\log_{4}(2016!)^{4}}+\dots+\dfrac{2016\times2016}{\log_{2016}(2016!)^{2016}}

S = 2016 ( 2 2 log 2 2016 ! + 3 3 log 3 2016 ! + 4 4 log 4 2016 ! + + 2016 2016 log 2016 2016 ! ) S=2016(\dfrac{2}{2\log_{2}2016!}+\dfrac{3}{3\log_{3}2016!}+\dfrac{4}{4\log_{4}2016!}+\dots+\dfrac{2016}{2016\log_{2016}2016!})

S = 2016 ( 1 log 2 2016 ! + 1 log 3 2016 ! + 1 log 4 2016 ! + + 1 log 2016 2016 ! ) S=2016(\dfrac{1}{\log_{2}2016!}+\dfrac{1}{\log_{3}2016!}+\dfrac{1}{\log_{4}2016!}+\dots+\dfrac{1}{\log_{2016}2016!})

S = 2016 ( log 2 log 2016 ! + log 3 log 2016 ! + log 4 log 2016 ! + + log 2016 log 2016 ! ) S=2016(\dfrac{\log2}{\log2016!}+\dfrac{\log3}{\log2016!}+\dfrac{\log4}{\log2016!}+\dots+\dfrac{\log2016}{\log2016!})

S = 2016 ( log 2 + log 3 + log 4 + + log 2016 log 2016 ! ) S=2016(\dfrac{\log2+\log3+\log4+\dots+\log2016}{\log2016!})

S = 2016 ( log ( 2 × 3 × 4 × × 2016 ) log 2016 ! ) S=2016(\dfrac{\log(2\times3\times4\times\dots\times2016)}{\log2016!})

S = 2016 ( log 2016 ! log 2016 ! ) S=2016(\dfrac{\log2016!}{\log2016!})

S = 2016 ( 1 ) S=2016(1)

S = 2016 S=2016

Great!, same solution :) +1

Novril Razenda - 4 years, 11 months ago

S = n = 2 2016 2016 n l o g n ( 2016 ! ) n = n = 2 2016 2016 n l o g 2016 ! ( n ) n = 2016 l o g 2016 ! ( 2016 ! ) = 2016 S=\displaystyle \sum_{n=2}^{2016} \frac{2016n}{log_{n}(2016!)^n} = \sum_{n=2}^{2016} \frac{2016n \cdot log_{2016!}(n)}{n} =2016\cdot log_{2016!}(2016!)=2016

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