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Cryptarithmetic Multiplication 31


1. Value of T ? 
(a) 3 (b) 4 (c) 5 (d) 6
2. Value of P? 
(a) 5 (b) 1 (c) 3 (d) 4
3. Value of K 
(a) 2 (b) 8 (c) 0 (d) 1

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      1. Kanchana

        So when you start replacing with the T=4, you will find that Q gets the value 4, and thats not possible! Cause T is already 4.

    1. Srujan K'mat

      Because there is already ‘2’ mentioned in the question. So W can’t be ‘2’.
      Also in the solution, I guess ‘2’ shouldn’t be there in the possible values of ‘W’.

      1. Kanchana

        I dont think taking W = 2 is an issue because 2 is already given. i replaced the values in case 5 in the main question, and found that Q is getting the value 4, not possible. Therefore move to next case.

  1. vikas solanki

    one more way to solve this question.
    we have possible values of
    p= {5,1,3,4,} and
    k = {2,8,0,1}
    and if you will look closely you will find value of p cant be 1
    and value of k cant be 0 and 1 (according to basic crypt rule)
    so only possible set of values for p and k are
    possible set of p and k =
    { (5,2),(5,8),(3,2),(3,8),(4,2),(4,8)}
    so start hit and trial with these 6 cases and you will get the answer with second element i.e.. p=5 and k=8.

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