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For all $n \in \mathbb{N}$ and prime $p$ i was asked to prove that $p$ divides $\binom{n}{p}$ -$[n/p]$. Where $[\text{...}]$ denotes box function and $\binom{n}{p}$ denotes $n$ choose $p$.

Can not understand where to start from. Is it possible by congruency? Please help.

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    What is $nCp$?? Is it $\binom{n}{p}$? – 2017-01-11
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    @DietrichBurde yes. I'm not sure what the "box function" is though. – 2017-01-11
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    Yes. N choose p i.e. n!/(p!×(n-p)!). – 2017-01-11
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    Box function is the greatest integer function. i.e. greatest integer less then or equal to the given value or parameter – 2017-01-11

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By Lucas's theorem $\binom{n}{p}$ is congruent to $\binom{w}{1}$ where $w$ is the second digit in the base $p$ expansion of $b$ ( because the base $p$ expansion of $p$ is just $10$). So $\binom{n}{p}$ is congruent to $w$ and $w$ is clearly congruent to $\lfloor n/p\rfloor$

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    Actually i dont know Lucas' theorem. I am in class 11. Can u explain a bit more clearly? – 2017-01-11
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    https://en.wikipedia.org/wiki/Lucas's_theorem ${}{}{}{}$ – 2017-01-11
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    Lucas's theorem is more general than what is needed for this problem, and a student at grade 11 would have to prove Lucas' theorem first to then prove this problem; that is probably not the easiest way to do it. – 2017-01-11
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    why the downvote? Has anyone seen a proof of Luca's theorem? It is direct and does not pass through the proof of this problem at all. – 2017-01-13