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Cryptarithmetic Problem 4

Cryptarithmetic Multiplication_04

1. Value of W + H + Y ? 
(a) 7 (b) 8 (c) 9 (d) 10
2. Value of B ? 
(a) 1 (b) 2 (c) 3 (d) 4
3. Value of N + U + T ? 
(a) 17 (b) 18 (c) 19 (d) 20

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11 comments

    1. rajan

      actually ,B + 6 = N , is not creating any carry otherwise ,at last B would not be B.
      So,if (B + 6 + (0 or 1 or 2) = N)
      then, N < 9 ,so possible values of, N={7, 8, 9}
      now if carry =0 ; B ={ 1, 2 ,3}
      else if carry = 1;B = { 0 , 1, 2}
      else if carry = 2 ;B={ 0 ,1} and N = { 8, 9}

    1. ElitmusZone Post author

      In the solution of the problem we taken two cases (Case1 and Case2) at the beginning of the solution.
      Case-1 When Y={2, 4, 8} and U={6}
      Case-2 When Y={5} and U={3, 7, 9}

      We have to do hit and trial with the both the cases. You can also take Case1 and proceed. Later on, Y=5 will get rejected.

      Please try once with Y=5.
      (We haven’t made any assumption while providing solution to this Cryptarithmetic Problem.)
      Please let us know if more information is required.

    2. Amarpal Singh

      We have U*Y=Y
      if we take u=5
      then
      5*1=5(not 1),5*2=0(not 2),5*3=5(not 3)……………means right digit never come
      but in case of u=6
      6*2=12(means 2)
      6*4=24(means 4)
      6*8=48(means 8)

  1. karan

    these type of questions can only be solved by trial and error method .you have to take values from options to get a answer otherwise you will get stuck in those problems.

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