Let us learn a quick method to multiply two numbers when one of the factors of the product is of the form \(11\), \(101\), \(1001\)\(...\) etc., while the other factor is any number with arbitrary number of digits.
This procedure enables us to obtain the product in a single step without employing a long multiplication.
Let us consider a four digit number \(dcba\), Where.
- \(d\) - represents the number in thousands place.
- \(c\) - represents the number in hundreds place.
- \(b\) - represents the number in tens place.
- \(a\) - represents the number in ones place.
The product of \(dcba\) with \(11\) using distributive property is given as follows:
\(dcba\) \(\times\) \(11\) \(=\) \(dcba\)\((10+1)\)
\(=\) \(dcba \times 10\) \(+\) \(dcba\)
\(=\) \(dcba0\) \(+\) \(dcba\)
The sum is represented in column wise as follows:
\(\begin{array}{cccccc}
& d & c & b & a & 0 \\
+ & & d & c & b & a \\
\hline
& d & (c+d) & (b+c) & (a+b) & a \\
\end{array}\)
& d & c & b & a & 0 \\
+ & & d & c & b & a \\
\hline
& d & (c+d) & (b+c) & (a+b) & a \\
\end{array}\)
From the above representation it is observed that, the first and the last digits of the number remain constant, while the in between numbers are the sum of the adjacent digits in the number.
Example:
Consider the product \(3678 \times 11\).
\(3678\) \(\times\) \(11\) \(=\) \(3678\)\((10+1)\)
\(=\) \(3678 \times 10\) \(+\) \(3678\)
\(=\) \(36780\) \(+\) \(3678\)
The sum is represented in column wise as follows:
\(\begin{array}{cccccc}
& (1) & (1) & (1) & & \\
& 3 & 6 & 7 & 8 & 0 \\
+ & & 3 & 6 & 7 & 8 \\
\hline
& 4 & 0 & 4 & 5 & 8 \\
\end{array}\)
& (1) & (1) & (1) & & \\
& 3 & 6 & 7 & 8 & 0 \\
+ & & 3 & 6 & 7 & 8 \\
\hline
& 4 & 0 & 4 & 5 & 8 \\
\end{array}\)
Thus, the product \(3678 \times 11\) \(=\) \(40458\).
Notice that, the sum of the adjacent numbers in between are added using standard addition rule. If the sum is \(\leq9\), then it is written as it is; if the sum is \(\geq10\), the number in the tens place is carried over to the next column.
Procedure to write the product of a number with \(11\) in a single step:
Based on the above generalisation and the example, the procedure to write the product of a number with \(11\) in a single step are as follows:
Step -1: Begin with the last digit of the number. Write it down as it is.
Step - 2: From right to left, add each pair of adjacent digits in the number (or add each number with the next number) and write the sum in between. If the sum is \(\geq 10\), carry over the tens digit to the next place.
Step - 3: Finally, write down the initial digit of the number, with any carry over if applicable.