Inspired by Xuming Liang

Algebra Level 4

Compute ⌊ π ⌋ + ⌊ π + 1 900 ⌋ + ⌊ π + 2 900 ⌋ + ⋯ + ⌊ π + 899 900 ⌋ . \left\lfloor \pi\right \rfloor + \left\lfloor \pi +\frac{1}{900} \right\rfloor +\left \lfloor \pi + \frac{2}{900} \right\rfloor +\cdots + \left\lfloor \pi + \frac{899}{900} \right\rfloor .


The answer is 2827.

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Alan Yan by Alan Yan , 20, USA

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1 solution

Alan Yan
Nov 27, 2015

By Hermite's Identity, ⌊ π ⌋ + ⌊ π + 1 900 ⌋ + ⌊ π + 2 900 ⌋ + . . . + + ⌊ π + 899 900 ⌋ = ⌊ 900 π ⌋ = 2827 \lfloor \pi \rfloor + \lfloor \pi + \frac{1}{900} \rfloor + \lfloor \pi + \frac{2}{900} \rfloor + ... + + \lfloor \pi + \frac{899}{900} \rfloor = \lfloor 900 \pi \rfloor = 2827

8 Helpful 0 Interesting 0 Brilliant 0 Confused

This way is so amazing for me. (I can solve this problem by the ordinary way.)

Panya Chunnanonda - 5 years, 5 months ago

Can we solve it in any other way,without using the identity.I guess we can use progressions but it becomes too complicated to deal with π. Do you have any simpler method (without using identity).

Adarsh pankaj - 5 years, 5 months ago

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Note that ⌊ π + k 900 ⌋ \left \lfloor \pi + \frac{k}{900} \right \rfloor can only be 3 3 or 4 4 . Thus, just find the k k such that π + k 900 ≥ 4 \pi + \frac{k}{900} \geq 4 and it is just a matter of summing terms.

Alan Yan - 5 years, 5 months ago

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