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Home/ Questions/Q 404943
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T17:21:16+00:00 2026-05-12T17:21:16+00:00

Here is a different approach for the Project Euler #1 solution: +/~.(3*i.>.1000%3),5*i.>.1000%5 How to

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Here is a different approach for the Project Euler #1 solution:

+/~.(3*i.>.1000%3),5*i.>.1000%5

How to refactor it?

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  1. Editorial Team
    Editorial Team
    2026-05-12T17:21:16+00:00Added an answer on May 12, 2026 at 5:21 pm
    [:+/@~.@,3 5([*i.@>.@%~)]
    

    usage example:

    f =: [:+/@~.@,3 5([*i.@>.@%~)]
    f 1000
    

    or

    +/~.,3 5([*i.@>.@%~)1000
    

    %~                        = 4 : 'y % x'
    i.@>.@%~                  = 4 : 'i. >. y % x'
    [*i.@>.@%~                = 4 : 'x * i. >. y % x'
    3 5([*i.@>.@%~)]          = 3 : '3 5 * i. >. y % 3 5'
    [:+/@~.@,3 5([*i.@>.@%~)] = 3 : '+/ ~. , 3 5 * i. >. y % 3 5'
    
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