Theory:

Like how dealing with numbers could be a fascinating experience, the mix of numbers and the alphabet letters could also be equally fascinating.
 
Let us consider the following puzzle.
 
Instructions:
 
1. Each of the letters of the alphabet will only constitute one-digit.
 
2. The first digit of a number can never be '\(0\)'.
 
Let us try to solve the following puzzle.
 
1320_1.svg
 
In this puzzle, we should try to find the value of \(A\).
 
Let us consider the entries in the ONES column.
 
\(A + 4 = 1\)
 
It is now understood that \(A\), when added with \(4\), gives a number ending with \(1\).
 
What are the numbers that end with \(1\)?
 
\(1\), \(11\), \(21\), \(31\),\(...\)
 
Let us look at the trial and error method:
 
Variable
\(A\)
Operation
\(A + 4\)
Final answer
\(0\)
\(0 + 4\)
\(4\)
\(1\)
\(1 + 4\)
\(5\)
\(2\)
\(2 + 4\)
\(6\)
\(3\)
\(3 + 4\)
\(7\)
\(4\)
\(4 + 4\)
\(8\)
\(5\)
\(5 + 4\)
\(9\)
\(6\)
\(6 + 4\)
\(10\)
\(7\)
\(7 + 4\)
\(11\)
 
When we consider \(A\) as \(7\), the final value is a number ending with \(1\).
 
Let us consider the HUNDREDS column.
 
\(1 + A = 8\)
 
Let us check if the condition \(A = 7\) is satisfied or not.
 
\(1 + 7 = 8\)
 
The condition is thus satisfied.
 
If we substitute \(7\) in the place of \(A\), we get the following.
 
1320_1_1.svg
 
This type of puzzle solving is called 'Cracking of codes' or 'Cryptarithms'.