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ElitmusZone » How To Solve Cryptarithmetic Problems 03

# How To Solve Cryptarithmetic Problems 03

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    (consider last digit)
5 x 7 = _ 5    (consider last digit)
5 x 9 = _ 5    (consider last digit)

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

Complete Solution

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}

```

1. Yamini Singh

Nice Work..!Thanks Keep it up!

2. Name

why 1 has not been mentioned in case of 5

1. Abhishek

Since the initial expression is AxB=_A and not AxB=A, we are looking for 2 digit product such that the unit’s digit (A) is same as the first multiplier (A).

3. VIKASH

why can’t we take 4*6 = 24 as E*P = E