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