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    • Colourful numbers

      Three-digit numbers

      1 May, 1998
      May 1998

      A three-digit number is such that its second digit is the sum of its first and third digits.

      Prove that the number must be divisible by 11.

      Solution

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      Anonymous

      13 June 2011

      Permalink
      Comment

      All right, but how can you prove that a number like 62345678987654555548 (or even a bigger number) is divisible bij 11?

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      Anonymous

      18 August 2011

      In reply to three-digit numbers by Anonymous

      Permalink
      Comment

      yes.
      1 way: Use calculator for windows 7, which is totally capable of numbers larger than that.(12 digits longer than 62345678987654555548)
      the other speedy way:
      What is:
      6 - 2 + 3 - 4 ...?
      If it is 0 or 11 or -11, it is divisible...
      if it isn't, no.

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      Anonymous

      21 November 2013

      In reply to yes... by Anonymous

      Permalink
      Comment

      It can be done with 121 1+1=2 and 121 is divisble by 11

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      Anonymous

      30 August 2014

      In reply to three-digit numbers by Anonymous

      Permalink
      Comment

      You can, in this way:

      First (from the first digit), divide the no: into groups of 2.

      Eg: 62 34 56 78 98 76 54 55 55 48

      Then, add them up.

      Eg: 62 + 34 + 56 + 78 + 98 + 76 + 54 + 55 + 55 + 48 = 616

      If the sum is more than 100, repeat step 1 and 2.

      Eg: 6 16
      6 + 16 = 22

      If the sum obtained is divisible by 11, then the initial no: is divisible by 11.

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      Anonymous

      13 July 2011

      Permalink
      Comment

      Let digits are a,b,c
      Let b=a+c because second digit is sum of last and first
      So number=a*100+b*10+c*1
      =a*100+(a+c)*10+c*1
      =a*100+a*10+c*10+c*1
      =110*a+11*c
      =11*10*a+11*c
      =11*(10*a+c)
      That is the number is multiple of 11 so it is divisible by 11

      opksalu@gmail.com

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