Cryptarithmetic Problem 1

solution:

                 A  P  D
                 x  A  D 
              R  P  A  D
           D  D  C  D   
           D  P  C  E  D
As, P + C= _C
Hence value of P=9   Rule 1 - Case-II
Put P=9 and rewrite the problem,
                 A  9  D
                 x  A  D
              R  9  A  D
           D  D  C  D   
           D  9  C  E  D
further, you can see
                 A  9  D
                 x  A  D
              R  9  A  D
           D  D  C  D     
           D  9  C  E  D
Here D x D = _ D [ R  9  A  D]
Hence possible values of D={5, 6}  Detailed Explanation- Rule 2

Firstly take D=5 and rewrite the problem
                A  9  5
                x  A  5
             R  9  A  5
          5  5  C  5   
          5  9  C  E  5

                A  9  5
                x  A  5
             R  9  A  5
          5  5  C  5   
          5  9  C  E  5
Here, you can easily predict the value of R=3
So, the problem reduces to
                A  9  5
                x  A  5
             3  9  A  5
          5  5  C  5   
          5  9  C  E  5
As, A x 5 = _5 [ 5  5 C 5]
Hence possible values of A={3, 7, 9}    Detailed Explanation
and as you have already taken R=3, Hence A cannot be equal to 3.

[In Cryptarithmetic, each variable should have unique and distinct value]

Hence possible value of A={7, 9}
Now, start hit and trial with the possible values of A
Firstly take A=7
Put A=7, and  rewrite the problem again
              7  9  5
              x  7  5
           3  9  7  5
        5  5  C  5   
        5  9  C  E  5
Now you can easily predict the value of C and E.
              7  9  5
              x  7  5
           3  9  7  5
        5  5  6  5   
        5  9  6  2  5