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Cryptarithmetic Multiplication 31

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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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4 comments

  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
    p={5,3,4}
    k={2,8,1}
    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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