## Saturday, 12 November 2016

### Cryptographical Tutorial for eLitmus

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.
Example of Cryptarithmetic Problem
Detailed Explanation
 1 Suppose if you are considering  A=2, then  other variable in problem cannot have value equal to 2.i.e.  In the given problem above, B≠2, M≠2, R≠2, Y≠2 etc. 2 You cannot take someplace A=2 and someplace A=3 in a single problem.i.e. If you are getting  A=2 and A=3 in same problem, solution is wrong. 3 Numbers must not begin with zero.i.e. In the given problem above, Value of M≠0, B≠0.
Fundamental 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}
```

Fundamental Rules
```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}
```