Tan on the floor!

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A positive integer n ∈ S n \in S , the set of tanny integers, if and only if for some value of θ ∈ ( 0 , π 2 ) \theta \in (0,\frac{\pi}{2}) , ⌊ tan ⁡ 2 θ ⌋ + tan ⁡ θ = n \lfloor \tan^2{\theta} \rfloor + \tan{\theta} = n . When the elements of S S are arranged in increasing order, let N T 2014 N_{T_{2014}} denote the 2014 2014 th tanny integer. Find the last three digits of N T 2014 N_{T_{2014}} .


The answer is 210.

A Brilliant Member by A Brilliant Member , 24, India

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1 solution

Note that in the equation ⌊ tan ⁡ 2 θ ⌋ + tan ⁡ θ = n \lfloor \tan^2{\theta} \rfloor + \tan{\theta} = n , since ⌊ tan ⁡ 2 θ ⌋ \lfloor \tan^2{\theta} \rfloor & n n are integers, tan ⁡ θ \tan{\theta} must also be an integer. Since tan ⁡ x \tan{x} is a surjective function, let tan ⁡ θ = k ∈ N \tan{\theta} = k \in \mathbb{N} . This gives n = k 2 + k = k ( k + 1 ) n=k^2+k = k(k+1) that is, a product of two consecutive integers. This means S = { k ( k + 1 ) } k = 1 ∞ ⇒ N T 2014 = 2014 ⋅ 2015 = 4058210 S = \{k(k+1)\}_{k=1}^{\infty} \Rightarrow N_{T_{2014}} = 2014 \cdot 2015 = \boxed{4058210} . ■ \blacksquare

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Nice, done it in the same way.

Jit Ganguly - 7 years, 4 months ago

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