Cryptarithmetic Problem 4

Solution:

Firstly, you have to divide the problem in three parts, so that it will help you in collecting more clues.
(1)          T  H  E 
                x  N
          S  N  T  I
(2)          T  H  E 
                x  E
          P  I  A  E

(3)          T  H  E 
                x  P
          H  B  N  E
(Choose one among the three which has maximum number of clues.)
In this case, you can take case(2)
Here, E x E = _E
Therefore, possible values of E = {5, 6} Rule 2
As,
5 x 5 = _5 [25] (last digit)
6 x 6 = _6 [36] (last digit)
                  T  H  E  
                x P  E  N
               S  N  T  I
            P  I  A  E
         H  B  N  E      
         S  H  A  A  H  I
Further, we have one more clue E x P = _E
Hence, possible values of E and P are as follows.
Case I   -   When E=5 and  P={3, 7, 9}
Case II  -   When P=6 and  E={2, 4, 8}   Rule 3
Now, you have to start hit and trial with both the possible cases.
Firstly, take E=5 and P = {3, 7, 9}
Put E=5 and rewrite the problem again.
               T  H  5 
            x  P  5  N
            S  N  T  I
         P  I  A  5
      H  B  N  5      
      S  H  A  A  H  I
Further,  5 x N = _I [ S  N  T  I ]
[If you multiply 5 to any number, you will only get [0, 5] as their last digit.](5 x even =_0 and 5 x odd=_5)
Therefore, value of I = 0
Hence, possible value of N = {2, 4 ,6 ,8}
Now, E=5 and I=0 and write the problem again.
              T  H  5
           x  P  5  N
           S  N  T  0
        P  0  A  5
     H  B  N  5      
     S  H  A  A  H  0
Now, again divide the problem in three parts
(1)           T  H  5
                  x N
           S  N  T  O
 (2)          T  H  5
                  x 5
           P  0  A  5
 (3)          T  H  5
                  x P
           H  B  N  5
Take Case (2) as it has less number of variable in comparison to case (1) and Case(2)
 
  (2)       T  H  5
                x 5
         P  0  A  5
Earlier, you have only three possible values of P= {3, 7, 9}
you have to start hit and trial with the values of P
Firstly, take P=3
  (2)       T  H  5
                x 5
         3  0  A  5
Then T=6 [6 x 5 = 30] 
            T  H  5
          x P  5  N
         S  N  T  0
      P  0  A  5
   H  B  N  5       
   S  H  A  A  H  0
as T + 5 = H i.e. 6 + 5 =_1 [last digit]  Hence H=1, 
 
  (2)       6  1  5
                x 5
         3  0  A  5 [ 3  0  7  5]
If you compare side by side then you will get A=7
Put these values in the main problem,
T=6, H=1, E=5, P=3, I=0, A=7
Hence,
            6  1  5
         x  3  5  N
         S  N  6  0
      3  0  7  5
   1  B  N  5      
   S  1  7  7  1  0

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