# Simple ceiling function problem [closed]

Prove that $\lceil4n/3\rceil\le 4\lceil n/3\rceil$ for all integers $n$. Try to generalize this result to something where something other than 4 and 3 are used.

-

## closed as not constructive by Andrés E. Caicedo, The Chaz 2.0, Leonid Kovalev, Asaf Karagila, t.b.Aug 16 '12 at 12:42

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

Quoting a homework problem without adding anything of your own does not constitute a question. – Henning Makholm Apr 4 '12 at 16:43

Hint: you might think about the fact that all integers can be expressed as either $3k, 3k+1$, or $3k+2$
Hint: $\lceil n+n/3\rceil=n+\lceil n/3\rceil$