PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free DemoLike 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.
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.
This type of puzzle solving is called 'Cracking of codes' or 'Cryptarithms'.
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