## Saturday, 12 November 2016

### Solving Cryptarithmetic by Unit Method

Unit Digit Method (Hit and trial approach) applies to those Cryptarithmetic  Problems, where you don’t have even a single clues to start solving the cryptarithmetic problem. In this case you have to start hit and trial with the possible values of unit digit of the multiplication problem.
Solving a Cryptarithmetic Problem by using this method will take some time but, after applying all the cases of  this method,  you will definitely reach to the solution. It will take nearly 10-12 minutes to solve the problem.
Before you study Unit Digit Method, you should have a basic understanding of how to solve the Cryptarithmetic problems.
Unit  Digit Method
Step-1
Firstly divide the cryptarithmetic problem into three parts. If it is a 3 x 3 Cryptarithmetic Problem, then you have to convert it into 3 x 1 problem.
Step-2
After dividing the problem in three parts, analyse all the parts very closely and try to collect some clue.
Now, you have to choose one among 3 which has maximum number of clues.
Step-3
You have to start hit and trial with the possible values of  the variable which is present at the unit digit. {0,  1,  2,  3,  4,  5,  6,  7,  8,  9}
Step -4
At each step, you have to check, whether the values  satisfies  Basic Cryptarithmetic Rule
Example

```                       W  B  A
x  B  P  W
C  X  R  F
F  X  A  X
A  A  C  C
A  P  C  A  B  F
Firstly divide the problem in three parts. As,

(1)      W  B  A             (2)     W  B  A        (3)      W  B  A
x  W                        x  P                    x  B
C  X  R  F                  F  X  A  X              A  A  C  C

Now, you have to select one from three which has maximum number of clues.i.e. maximum number of variables getting repeated.
In this case take,
(3)          W  B  A
x  B
A  A  C  C
Now, you have to start hit and trial with the possible values C i.e.

C={1, 2, 3, 4, 5, 6, 7, 8, 9}

Firstly take  C=1 and check further

Possible ways of getting 1 as Unit Digit

(3)                                  W  B  A
x  B
A  A  C  C

then
Case 1     C=1  A=1  B=1   Rejected As A = B = C [violates the Basic Cryptarithmetic Rules.]
Case 2     C=1  A=7  B=3   Needs to be checked further
Case 3     C=1  A=3  B=7   Needs to be checked further
Case 4     C=1  A=9  B=9   Rejected As A = B  [violates the basic Cryptarithmetic Rules.]
Now check for Case 2 and Case 3 only.
(3)     W  B  A
x B
A  A  C  C

W  3  7
x  3
7  7  1  1
Rejected as,  even after taking the value of W=9, You will never get 77.
W  3  7
x  3
7  7  1  1
Now, you have to check with case 3
C=1, A=3, B=7
W  B  A
x  B
A  A  C  C

W  7  3
x  7
3  3  1  1
Here you can easily predict the value of W=4
Hence W=4, C=1, A=3, B=7 put these values in main problem
4  7  3
x  7  P  4
1  X  R  F
F  X  3  X
3  3  1  1
3  P  1  3  7  F
Now, you can see
4  7  3
x  4
1  8  9  2 [ 1  X  R  F]
If you compare side by side you will get X=8, R=9 and F=2
Put these values in main problem, and rewrite again
4  7  3
x  7  P  4
1  8  9  2
2  8  3  8
3  3  1  1
3  P  1  3  7  2
Now, you can easily predict the value of P=6
Hence,
4  7  3
x  7  6  4
1  8  9  2
2  8  3  8
3  3  1  1
3  6  1  3  7  2
[Relax - it's going to take some time to understand the whole concept. If you are facing any difficulty. Please go through the Cryptarithmetic Tutorial.]
```