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Home/ Questions/Q 6372951
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T01:18:12+00:00 2026-05-25T01:18:12+00:00

The following function counts how often I can divide one number by another: divs

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The following function counts how often I can divide one number by another:

divs n p = if (n `mod` p == 0) then 1 + divs (n `div` p) p else 0

Is there a shorter way to write divs?

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  1. Editorial Team
    Editorial Team
    2026-05-25T01:18:13+00:00Added an answer on May 25, 2026 at 1:18 am
    divs n p = case n `divMod` p of (q,0) -> 1 + divs q p; _ -> 0
    

    And more efficient too! Note that quotRem, which behaves differently with negative values, is a tiny-tiny bit more efficient still.

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