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

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

Cryptarithmetic Problem 2

Solution:

                 V  I  A
              x  G  O  T
              G  R  O  T
           A  A  R  O
        A  I  A  G      
        A  S  T  A  R  T
Here, 
A x T = _T [G R O T]
A x O = _0 [A A R O]
A x G = _G [A I A G]
This is only possible when A=1
Hence, Put A=1 and rewrite the problem again.

               V  I  1
            x  G  O  T
            G  R  O  T
         1  1  R  O
      1  I  1  G      
      1  S  T  1  R  T
At this stage, divide the problem in 3 parts for collecting more clues..
(1)            V  I  1
                  x  T 
            G  R  O  T
(2)            V  I  1
                  x  O
            1  1  R  O
(3)            V  I  1
                  x  G
            1  I  1  G
Now you have to choose one among three based on number of clues.
In this problem, you can take case (3).
(You can also take case (1) and Case (2))
(3)           V  I  1
                 x  G
           1  I  1  G
At this stage you have to start hit and trial with the possible values of G
G = {2, 3, 4, 5, 6, 7, 8, 9}
*G ≠ {1}
(As you have  already taken A=1)
[In Cryptarithmetic, each variable should have and unique and distinct value.]
*G ≠ {0}
(As you are multiplying some number by G in (3). If we take G=0)
then            V  I  1
                    x G
             0  0  0  0 [1  I  1  G]
Now,take G=2
(3)            V  I  1 
                   x 2
            1  I  1  2
*Rejected (you can see 2 x I =_1)

( You will never get unit digit at 1 after multiplying any number by 2.) 
If we multiply any digit by 2 we cannot get last digit as 1.
i.e. 2x1=2, 2x3=6, 2x4=8 2x5=_0, 2x6=_2, 2x8=_6[16] and 2x9=_8[18])

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

Now,take G=3
put G=3 in case(3)
(3)            V  I  1 
                   x 3
            1  I  1  3
Now we can  see I x 3 = _1 [ 1 I 1 3]
(last digit is 1 which is only possible when I=7 (7 x 3 = 21)
Put I=7 in (3)
(3)            V  7  1 
                   x 3
            1  7  1  3
Now, you can easily predict the value of V=5 (3 x 5 = 15 + 2(carry)=17)

(3)           5  7  1 
                 x  3
           1  7  1  3
Therefore, 
V=5, I=7, A=1, G=3, Put these value in main problem and solve further.

              5  7  1
           x  3  O  T
           3  R  O  T
        1  1  R  O
     1  7  1  3      
     1  S  T  1  R  T
Now you can easily predict the other values.
S=8, T=6, R=4, O=2
These values also satisfies the Basic Cryptarithmetic Rules

              5  7  1 
           x  3  2  6
           3  4  2  6
        1  1  4  2
     1  7  1  3      
     1  8  6  1  4  6

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     

Cryptarithmetic rules II

Rule-3
If  A x B = _ A then possible values of A and B
Case I 
When A = 5 and B = {3, 7, 9}
A x B = _ A
5 x 3 = _ 5  [15]  (consider last digit)
5 x 7 = _ 5  [35]  (consider last digit)
5 x 9 = _ 5  [45]  (consider last digit)

Case II 
When A = {2, 4, 8} and B = {6}
A x B = _ A
2 x 6 = _ 2  [12]  (consider last digit)
4 x 6 = _ 4  [24]  (consider last digit)
8 x 6 = _ 8  [48]  (consider last digit)

Example Supporting Case-I and Case-II
               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

                 
In this problem, P x E= _E [H  B  N  E]

Case 1   E={5} and P={3, 7, 9}

Case 2   P={6} and E={2, 4, 8}

Cryptarithmetic rules

Rule-1
If  A + B = A  then the possible value of B={0, 9}
Case I 
i.e. K + A = A (K=0)
Example Supporting Case-I 
                      P  A  S
                   x  R  B  Q
                   S  B  K  W
                A  S  A  A 
             S  E  P  B      
             S  Q  S  K  A  W
Here, you can easily predict the value of K = 0

Case II 

If A + B = A
When B = 9  9 + A = A   [9 + A + 1(carry) = _A]
When you have 1 carry from the previous addition.
Example supporting Case-II
                                                3  5 
                                               +9  7
                                             1  3  2
Example:

                      A  I  D
                      x  A  D
                   R  I  A  D
                D  D  C  D      
                D  I  C  E  D

Here, you can take I=9 as I + C = C

Rule-2
If A * A = _A  then the possible value of A={1, 5, 6}
Case I 
      1.    When A=1   1 * 1 = _1  

Case II
      1.    When A=5   5 * 5 = _5 (25) [consider last digit only]
      2.    When A=6   6 * 6 = _6 (36) [consider last digit only]

Example supporting Case-II
 
                       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


In this problem, you can easily predict value of A={5, 6}

Introduction to Cryptarithmetic

Fundamental Rules:

1.	Each Variable should have unique and distinct value.
2.	Each Letter, Symbol represents only one digit throughout the problem.
3.	Numbers must not begin with zero  i.e.  0123 (wrong) , 123 (correct).
4.	You have to find the value of each letter in the Cryptarithmetic.
5.	There must be only one solution to the problem.
6.	The Numerical base, unless specifically stated , is 10.
7.	After replacing  letters by their digits, the resulting arithmetic operations must be correct.