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.
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 they are.
Step - 2: Moving from right to left, take the remaining digits of the number (starting from the \((n+1)\)-th digit from the right). Add each digit to the digit located exactly \(n\) places to its right. Write down the single-digit sum, and carry over any tens (\(\ge 10\)) to the next addition on the left.
Step - 3: Finally, write down the leftmost \(n\) digits of the original number at the front of your answer, adding any final carry-over if applicable.
Step - 2: Moving from right to left, take the remaining digits of the number (starting from the \((n+1)\)-th digit from the right). Add each digit to the digit located exactly \(n\) places to its right. Write down the single-digit sum, and carry over any tens (\(\ge 10\)) to the next addition on the left.
Step - 3: Finally, write down the leftmost \(n\) digits of the original number at the front of your answer, adding any final carry-over if applicable.
Important!
Multiplying by \(99\), \(999\),....
Use the fact \(99 = 100 -1\)
Use the fact \(99 = 100 -1\)
Fast Multiplication Summary:
| Multiply by | Think as | Quick Method | Example |
|---|---|---|---|
| \(11\) | \(10 + 1\) | Start from the right. Add each digit to the digit beside it. Write the answer from right to left. If the sum is more than \(9\), carry \(1\) to the next step. | \(5284 × 11 = 58124\) |
| \(101\) | \(100 + 1\) | Shift the number \(2\) places to the left, then add the original number. | \(4628 × 101 = 467428\) |
| \(1001\) | \(1000 + 1\) | Shift the number \(3\) places to the left, then add the original number. | \(7352 × 1001 = 7359352\) |
| \(10001\) | \(10000 + 1\) | Shift the number \(4\) places to the left, then add the original number. | \(8416 × 10001 = 84168416\) |
| \(10…01\) (General Rule) | \(10ⁿ + 1\) | Move the number \(n\) places left, then add the original number. | Example: \(×100001 →\) shift \(5\) places, then add. |
| \(99\) | \(100 − 1\) | Multiply by \(100\), then subtract the original number. | \(6825 × 99 = 675675\) |
| \(999\) | \(1000 − 1\) | Multiply by \(1000\), then subtract the original number. | \(5248 × 999 = 5242752\) |
| \(9999\) | \(10000 − 1\) | Multiply by \(10000\), then subtract the original number. | \(3145 × 9999 = 31446855\) |
| \(10…0 − 1\) (General Rule) | \(10ⁿ − 1\) | Multiply by \(10ⁿ\), then subtract the original number. | Example: \(×99999 →\) multiply by \(100000\), then subtract. |