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
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}

You might be interested in this article |

● Cryptarithmetic Tutorial 01 |

● Cryptarithmetic Tutorial 02 |

● Cryptarithmetic Multiplication Problems |

● Introduction to Cryptarithmetic |

● Logical Reasoning Problems |

Nice Work..!Thanks Keep it up!

WRONG

why 1 has not been mentioned in case of 5

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).

why can’t we take 4*6 = 24 as E*P = E