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Cryptarithmetic Problem- 5

Somdeb Burman | 14:20 | 9 comments

Multiplication





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9 comments:

  1. Shivani2 November 2015 at 23:09

    Why can E not be 5?

    ReplyDelete
  2. Unknown21 November 2015 at 07:45

    If A+E+carry =Q,without generating any carry over,then how can we conclude that E != 5.
    please reply.looking forward for ur ans

    ReplyDelete
  3. Unknown24 March 2016 at 07:34

    2+5+carry (2) =9 no carry over
    3+5+carry (1) =9 no carry over
    1+5+carry (2) =8 no carry over
    1+5+carry (1) =7 no carry over

    ReplyDelete
  4. Unknown24 March 2016 at 07:50

    As A can only be 2, 4, 6, 8 (since A+2B = 20 0r 10 from third column from left, K=0) Why cant A=2 and E =5

    ReplyDelete
  5. Anonymous25 July 2016 at 03:47

    The value of Q will be 7.

    ReplyDelete
  6. Unknown15 August 2016 at 20:24

    the value of Q will be 7

    ReplyDelete
  7. Anonymous18 August 2016 at 01:01

    The approach to find the value of p i.e 5 is not good since there are less than ten variables so p can have other value as well.

    ReplyDelete
  8. Unknown18 August 2016 at 01:06

    better approach is to first check the following multiplication

    PAS* B=ASAA
    try to find answer from here. If you could not i will help u.

    ReplyDelete
  9. Unknown29 August 2016 at 02:17

    here Q=7 as R&Q both cant be 6.

    ReplyDelete