Armstrong Number Program in Python || Python Program to check whether a number is Armstrong or not

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  • Опубликовано: 10 янв 2025

Комментарии • 12

  • @KeerthanaYelaboina-v4p
    @KeerthanaYelaboina-v4p 25 дней назад

    Thank you so much Andi for your neat and clean explanation thanks a lot 🙂👍

    • @SudhakarAtchala
      @SudhakarAtchala  19 дней назад

      welcome. Please like the videos, subscribe to the channel, and share it with your friends. Thanks in advance.

  • @minethsenaratne
    @minethsenaratne 2 года назад +1

    Thank you so much for the video!🙌🏼

    • @SudhakarAtchala
      @SudhakarAtchala  2 года назад

      Welcome. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.

  • @jaimingurjar3043
    @jaimingurjar3043 2 года назад

    Very nice Videos sir , i got to understand many concepts from your videos

    • @SudhakarAtchala
      @SudhakarAtchala  2 года назад

      Thanks. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.

  • @saiLakshmi.N
    @saiLakshmi.N Год назад

    Sir,What about 0?

    • @SudhakarAtchala
      @SudhakarAtchala  Год назад

      It's not a armstrong number. Plz subscribe to the channel and if possible share with your friends. Thanks in advance..

  • @rithvikaa2846
    @rithvikaa2846 7 месяцев назад

    As we did not specify n value 15 after the 2nd iteration how will n take valuen 15 and return 1 when n//10 is performed sir?

    • @SudhakarAtchala
      @SudhakarAtchala  7 месяцев назад

      // gives floor division i.e int, 153//10 means 15, 15//10 means 1,1//10 means 0.
      153%10 means 3, 15%10 means 5, 1%10 means 1. Plz subscribe to the channel and if possible share with your friends. Thanks in advance..

  • @angelch5565
    @angelch5565 3 года назад

    2digit number is Armstrong or not

    • @SudhakarAtchala
      @SudhakarAtchala  3 года назад

      Yes, no 2 digit number is armstrong. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.