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Wednesday, May 10, 2017

Recreatii Matematice 2/2016, Problem VII.208

Problem:
Prove that the number N=2016^{n+1}-2015n-2016 has at least 27 divisors for any n \in \mathbb{N}^*.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
We have \begin{array}{lll} N&=&2016\cdot(2016^n-1)-2015n\\&=&2016\cdot2015\cdot(2016^{n-1}+2016^{n-2}+\ldots+1)-2015n\\&=&2015\cdot(2016^n+2016^{n-1}+\ldots+2016-n)\\&=&2015\cdot[(2016^n-1)+(2016^{n-1}-1)+\ldots+(2016-1)] \end{array}
Since 2016^n-1 is divisible by 2015 for all n \geq 1, we have that 2015^2 \ | \ N. As 2015=5\cdot13\cdot31, then 2015=5^2\cdot13^2\cdot31^2, which has 27 divisors. The conclusion follows.

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