From floor to ceiling

Find a positive integer nn such that 20n13+13n20=2013 \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 x:RZ\lfloor x \rfloor: \mathbb{R} \rightarrow \mathbb{Z} refers to the greatest integer smaller than or equal to xx. For example 2.3=2\lfloor 2.3 \rfloor = 2 and 5=5\lfloor -5 \rfloor = -5.

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

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