Wednesday, 28 June 2017

Cryptarithmetic Problem 3

Cryptarithmetic Problem 3

Solution:

                 T  E  A
              x  H  A  D
              L  D  T  R
           H  R  S  A
        E  W  D  A      
        L  E  S  S  E  R
Here, A x A = _ A  [ H R S A ]
Therefore, Possible values of A= {5, 6}   Detailed Explanation- Rule 2
Further,
                 T  E  A
              x  H  A  D
              L  D  T  R
           H  R  S  A
        E  W  D  A     
        L  E  S  S  E  R
Here, H x A= _A [ E W D A ]
Therefore, two possible cases for the values of H and A
Case I  -  when A={5} then H={3, 7, 9}
Case II -  when A={2, 4, 8}  then H={6}    Detailed Explanation-Rule 3
Firstly taking case - I
Taking A=5 rewrite the problem again,
               T  E  5
            x  H  5  D
            L  D  T  R
         H  R  S  5
      E  W  D  5      
      L  E  S  S  E  R
Further,
               T  E  5
            x  H  5  D
            L  D  T  R
         H  R  S  5
      E  W  D  5      
      L  E  S  S  E  R
Here, 5 x D = _ R [ L D T R ]
Now, you can easily predict the value of R = 0 and possible values of D= {2, 4, 6, 8}
[If you multiply 5 to a number, you will only get[0,5] as their unit digit.]
5 x Even =_0   [2, 4, 6, 8]
5 x Odd  =_5   [3, 5, 7, 9]

[Relax it's going to take some to understand the concept. Please read... again!]

Put R=0 and write the problem again,
             T  E  5
          x  H  5  D
          L  D  T  0
       H  0  S  5
    E  W  D  5      
    L  E  S  S  E  0
At this stage, divide the problem into 3 parts,
(1)          T  E  5
                x  D
          L  D  T  0
(2)          T  E  5
                x  5
          H  0  S  5
(3)          T  E  5
                x  H
          E  W  D  5
Now, take (2) [ As it has less number of variables. 5 is repeated three times.]
(2)          T  E  5
                x  5
          H  0  S  5
Now you have to start hit and trial with the possible values of E
Firstly take E=1
Put E=1 in (2)
 (2)           T  1  5
                  x  5
            H  0  S  5 [ H  0  7  5]
If you compare side by side, then you will get S=7
Put S=7 and E=1 in the main problem.
[It needs to be checked further whether these values satisfies the Basic Cryptarithmetic Rules]
                T  1  5
             x  H  5  D
             L  D  T  0
          H  0  7  5
       1  W  D  5      
       L  1  7  7  1  0
At this stage you can easily predict all the values as
You can see T + 5 =_1 (which is only possible when the value of the T=6)
L=2 (As, 1 + 1(carry) = L)
Hence T=6, L=2.
Now you can easily solve the problem.
              6  1  5
           x  3  5  4
           2  4  6  0
        3  0  7  5
     1  8  4  5      
     2  1  7  7  1  0

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