Earlier, we learnt a quick method for multiplication using the distributive property when one of the factors of the product is of the form \(11\), while the other factor is any number with an arbitrary number of digits.
Now, let us extend the method to cases where one of the factors of the product is of the form \(101\), \(1001\), etc., while the other factor can again be any number with any arbitrary number of digits.
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.
Procedure to write the product of a number with \(101\) in a single step:
The product of \(dcba\) with \(101\) using distributive property is given as follows:
\(dcba\) \(\times\) \(101\) \(=\) \(dcba\)\((100+1)\)
\(=\) \(dcba \times 100\) \(+\) \(dcba\)
\(=\) \(dcba00\) \(+\) \(dcba\)
The sum is represented in column wise as follows:
\(\begin{array}{ccccccc}
& d & c & b & a & 0 & 0\\
+ & & & d & c & b & a \\
\hline
& d & c & (b+d) & (a+c) & b & a \\
\end{array}\)
& d & c & b & a & 0 & 0\\
+ & & & d & c & b & a \\
\hline
& d & c & (b+d) & (a+c) & b & a \\
\end{array}\)
From the above representation it is observed that, the first and the last two digits of the number remain constant, while the in between numbers are the sum of the alternate digits in the number.
Thus, the general rule to find the product are as follows:
Step -1: Begin with the last two digits of the number. Write it down as it is.
Step - 2: From right to left, add each pair of alternate digits in the number (or add each number with the second 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 two digits of the number, with any carry over if applicable.
Example:
Consider the product \(3678 \times 101\).
Step -1: Write the last two digits \(78\) as it is.
Step - 2: From right to left, add the alternate numbers.
\(6+8\) \(=\) \(14\)
So, write \(4\) and carry over \(1\) to the next sum.
Carry over \(1\) \(+\) \(3+7\) \(=\) \(11\)
So, write \(1\) and carry over \(1\) to the next sum.
Step - 3: Write the initial two numbers including the carry over.
\(36 +1\) \(=\) \(37\)
Thus, the product \(3678 \times 101\) \(=\) \(371478\).
Procedure to write the product of a number with \(1001\) in a single step:
The product of \(dcba\) with \(1001\) using distributive property is given as follows:
\(dcba\) \(\times\) \(1001\) \(=\) \(dcba\)\((1000+1)\)
\(=\) \(dcba \times 1000\) \(+\) \(dcba\)
\(=\) \(dcba000\) \(+\) \(dcba\)
The sum is represented in column wise as follows:
\(\begin{array}{cccccccc}
& d & c & b & a & 0 & 0 & 0\\
+ & & & & d & c & b & a \\
\hline
& d & c & b & (a+d) & c & b & a \\
\end{array}\)
& d & c & b & a & 0 & 0 & 0\\
+ & & & & d & c & b & a \\
\hline
& d & c & b & (a+d) & c & b & a \\
\end{array}\)
From the above representation it is observed that, the first and the last three digits of the number remain constant, while the in between numbers are obtained by adding each digits with the second next number.
Thus, the general rule to find the product are as follows:
Step -1: Begin with the last digits of the number. Write it down the last three digits as it is.
Step - 2: From right to left, add each number with the third 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 three digits of the number, with any carry over if applicable.
Example:
Consider the product \(3678 \times 1001\).
Step -1: Write the last three digits \(678\) as it is.
Step - 2: From right to left, add each number with the third next number.
\(8+3\) \(=\) \(11\)
So, write \(1\) and carry over \(1\) to the next sum.
Step - 3: Write the initial three numbers including the carry over.
\(367 +1\) \(=\) \(368\)
Thus, the product \(3678 \times 1001\) \(=\) \(3681678\).
Procedure to write the product of a number with \(10000...1\) in a single step:
In general, the method to find the product when one of the factors of the product is of the form \(1000...1 = 10^n+1\), while the other factor is any number with an arbitrary number of digits is as follows:
Step -1: Begin with the last digits of the number. Write it down the last \(n\) digits as it is.
Step - 2: From right to left, add each number with the \(n\)th 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 \(n\) digits of the number, with any carry over if applicable.
Step - 2: From right to left, add each number with the \(n\)th 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 \(n\) digits of the number, with any carry over if applicable.