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Thread: Cryptarithmetic 2

  1. #1
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    Cryptarithmetic 2

    Solve this,

    DONALD
    GERALD+
    --------
    ROBERT


  2. #2
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    Re: Cryptarithmetic 2

    1 4 2 5 2 1 3 0 4 5 2 1 + ----------- 4 4 7 0 4 2


  3. #3
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    Re: Cryptarithmetic 2

    526485
    197485
    ______
    723970

    This was a hard one! (For me )


  4. #4
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    Re: Cryptarithmetic 2

    Barbie ur answer was wrong.
    The condition was all letters have unique number.
    Check it in the equation.

    --------------------------
    suresh


  5. #5
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    Thumbs up Re: Cryptarithmetic 2

    (Well I tried many other ways too... This way yielded)
    1)After a lot of permutation and combinations, I assumed D= 5, then T = 0
    2)Now, O + E = O, then e Must be 9
    3) Also, L + L + 1 (1 is carry from adding D+D) = R, then R must be an odd number
    4)Now, R must be greater than 5 because 5 + G = R. Since E=9, R must be 7 as 7 is the only odd number between 5 and 9.
    5)Now, 5 + 1 + G = 7. So G will have to be 1.
    6)L+L+1=R. R=7. So, L is either 3 or 8.
    Now, A+A=E. But,E=9 which means, there has to be a carry.
    Which makes L=8
    7)N+7=B (There are 3 unused numbers now. 2,3 and 6)
    => N=6 and B=3
    8) O=2 (@ is the only left over number)

    Boy! Wasn't that hard?


  6. #6
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    Re: Cryptarithmetic 2

    Barbie, the rule says no two letters will have same digit assigned though not mentioned it is implicit. Guess you used same digit for many alphabets....


  7. #7
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    Re: Cryptarithmetic 2

    Ya.. I used so. I am not aware of that.
    Thank you.


  8. #8
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    Re: Cryptarithmetic 2

    That's no problem. Now you know ;-)


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