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Why is it that whenever you multiply any number by nine the digits of the result always add up to a multiple of nine? Example:

9 * 7 = 63;         // 6 + 3 = 9
9 * 35 = 315;         // 3 + 1 + 5 = 9

  • 0
    $11\cdot 9=99,9+9=18, 1111*9=9999,9+9+9+9=36?$ It should be mutiple of 9?2012-10-04
  • 0
    @labbhattacharjee The poster probably meant that you continue adding digits until you get one digit. 1+8=9 and 3+6=92012-10-04
  • 2
    Much more usefully, if the result of this process *does* produce $9$, then your original number is a multiple of $9$.2012-10-04
  • 2
    In all the proofs the crucial fact is that $10\equiv 1 \ \ (9)$.2012-10-04

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