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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.

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

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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 $3k+2$

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Hint: $\lceil n+n/3\rceil=n+\lceil n/3\rceil$

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