Divisible by this year? (Part 7)

Calculus Level 4

Let $n$ and $k$ be positive integers such that the $\lim _{ x\rightarrow 0 }{ \left\lfloor nx\cot { x } \right\rfloor } = k$ . Find the minimum possible value of $n$ if $k$ is divisible by 2014.

This problem was inspired by a problem in Sipnayan 2014.

None of the above 2014 2013 2015

1 solution

Rindell Mabunga
Dec 13, 2014

Since $\lim _{ x\rightarrow 0 }{ \left\lfloor x\cot { x } \right\rfloor }$ always approach to $1^-$ , $\lim _{ x\rightarrow 0 }{ nx\cot { x } }$ will approach $n$ on the left side and thus by adding the floor function $\lim _{ x\rightarrow 0 }{ \left\lfloor x\cot { x } \right\rfloor } = n - 1$ .

The smallest positive integer $n$ that makes $n - 1$ divisible by $2014$ is $2015$ .

Yeah, I concluded that from the Taylor series of $\cot x$ . It's also easy to see from the graph of $x \cot x$ .

Jake Lai - 6 years, 6 months ago

This is a wrong solution. By your own argument if n=1 the limit equals n-1 = 1-1 = 0. It is known that 0 is divisible by all numbers, including 2014. So the answer should be 1 or, "none of the above".

Leonel Castillo - 3 years, 4 months ago

Divisible by this year? (Part 7)

This problem was inspired by a problem in Sipnayan 2014.

1 solution

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