From floor to ceiling

Find a positive integer \(n\) such that \( \left\lfloor \frac {20n}{13} \right\rfloor + \left\lceil \frac {13n}{20} \right\rceil = 2013 \).

This problem is proposed by Ahaan Rungta.

Details and assumptions:

  • The function \(\lfloor x \rfloor: \mathbb{R} \rightarrow \mathbb{Z}\) refers to the greatest integer smaller than or equal to \(x\). For example \(\lfloor 2.3 \rfloor = 2\) and \(\lfloor -5 \rfloor = -5\).

  • The function \( \lceil x \rceil : \mathbb{R} \rightarrow \mathbb{Z} \) refers to the smallest integer that is greater than or equal to to \(x\). For example, \( \lceil 2.3 \rceil = 3 \) and \( \lceil -5 \rceil = -5 \).

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