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Similar to the logics on two-digit numbers, three-digit numbers can also be subjected to a few tricks.
Trick \(1\): What happens to the difference between three-digit numbers and their reverse?
The difference between three digit numbers and their reverse is always a multiple of \(99\).
Consider the three-digit number '\(abc\)'.
The general form of '\(abc\)' is \((100 \times a) + (10 \times b) + c\) or \(100a + 10b + c\).
The reverse of '\(abc\)' is '\(cba\)'.
The general form of '\(cba\)' is \((100 \times c) + (10 \times b) + a\) or \(100c + 10b + a.\).
Now, let us find the difference between '\(abc\)' and '\(cba\)'.
If \(a > c\):
\(abc - cba = (100a + 10b + c) - (100c + 10b + a)\)
\(= 100a + 10b + c - 100c - 10b - a\)
\(= 99a - 99c\)
\(= 99(a - c)\)
If \(c > a\):
\(cba - abc = (100c + 10b + a) - (100a + 10b + c)\)
\(= 100c + 10b + a - 100c - 10b - a\)
\(= 99c - 99a\)
\(= 99(c - a)\)
Example:
Let us impose this logic on \(325\).
The reverse of \(325\) is \(523\).
Hence, to find the difference, we should subtract \(325\) from \(523\).
\(523 - 325 = 198\)
\(= 99 \times 2\)
Hence, it is proved that the difference between three-digit numbers and their reverse is a multiple of \(99\).
Trick \(2\): What happens when \(3\) forms of a three-digit number is summed up?
Consider the number '\(abc\)'.
Form \(1\): \(abc = 100a + 10b + c \longrightarrow (1)\)
To find the other forms of the number, shift the ONES digit to the number's left end.
Therefore, the number '\(abc\)' becomes '\(cab\)'.
The digit '\(c\)' is shifted to the left end of the number.
Form \(2\): \(cab = 100c + 10a + b \longrightarrow (2)\)
To form the third number, shift '\(b\)' from '\(cab\)' to the left end.
Form \(3\): \(bca = 100b + 10c + a \longrightarrow (3)\)
On adding \((1)\), \((2)\), and \((3)\), we get:
\(abc + cab + bca = 100c + 10a + b + 100b + 10c + a + 100b + 10c + a\)
\(= 111(a + b + c)\)
\(= 37 \times 3(a + b + c)\)
Therefore, the sum of \(3\) forms of a three-digit number is always a multiple of \(37\).
Example:
Let us consider the number \(128\).
Form \(1\): \(128\)
Form \(2\): \(812\)
Form \(3\): \(281\)
On adding the \(3\) numbers, we get:
\(128 + 812 + 281 = 1221\)
\(= 37 \times 33\)
Thus, the sum of three forms of a three-digit number is always a multiple of \(37\).
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