Not math, but perl -le 'print sprintf("%.0f", 5/3);'.

Try perldoc -q round for a more complete answer.

Update: monsieur_champs++ might well be right, but the AnonyMonk's question is unclear. 5/3 rounded to the nearest integer is 2, but so is 5 % 3 :-)

If the information in this post is inaccurate, or just plain wrong, don't just downvote - please post explaining what's wrong.
That way everyone learns.

In reply to Re: mathematical operators by BazB
in thread mathematical operator for integer division, "div" ? by Anonymous Monk

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