# Multiplying Whole Numbers

Let's start multiplying whole numbers.

Suppose 5 of your friends gave you 8 apples. How many apples do you have? It is easily to calculate it using addition: 8+8+8+8+8=40.

But it is very long and time consuming to make such calculations. Moreover, suppose, you have 100 friends that give you 5 apples. You need to add five 100 times!

For this we use multiplication.

Multiplication of numbers ${a}$ and ${b}$ is $\color{purple}{a \cdot b=\underbrace{a+a+a+a+...+a}_{b}}$.

For example, ${3}\cdot{4}={3}+{3}+{3}+{3}$, ${6}\cdot{7}={6}+{6}+{6}+{6}+{6}+{6}+{6}={42}$.

Multiplication tells us how many times to use number in addtion.

Each number being multiplied is called factor and the result is called product.

So, in ${3}\cdot{4}={12}$ both 3 and 4 are factors and 12 is product.

There is another notation for multiplication: $\times$, so ${a}\cdot{b}$ and ${a}\times{b}$ are equivalent.

A nice property of multiplication is that order doesn't matter.

For example ${3}\times{4}={3}+{3}+{3}+{3}={12}$ and ${4}\times{3}={4}+{4}+{4}={12}$. As can be seen result is same.

For any numbers ${a}$ and ${b}$ we have that ${\color{blue}{{{a}\times{b}={b}\times{a}}}}$.

Another interesting facts:

• Zero multiplied by any number is zero. For example, ${25}\times{0}={0}$, ${0}\times{1568}={0}$.
• Any number multiplied by 1 is number itself. For example, ${67}\times{1}={67}$, ${1}\times{8445}={8445}$.

For result of multiplication of some numbers see multiplication table.

Now, let's see how to multiply any whole numbers.

Example 1. Find ${34}\times{12}$.

Let's write numbers one under another:

$\begin{array}{l@{\,}l@{\,}l} \ & 3&4 \\ \times & \color{blue}{1}&\color{green}{2} \\ \hline & & \\ \end{array}$

Start with the right (green) digit. Multiply it with all digits of first number:

$\begin{array}{l@{\,}l@{\,}l} \ & \color{brown}{3}&\color{purple}{4} \\ \times & \color{blue}{1}&\color{green}{2} \\ \hline & \color{green}{2 \times \color{brown}{3}}&\color{green}{2 \times \color{purple}{4}} \\ \end{array}$

$\begin{array}{l@{\,}l@{\,}l} \ & \color{brown}{3}&\color{purple}{4} \\ \times & \color{blue}{1}&\color{green}{2} \\ \hline & 6&8 \\ \end{array}$

Now, multiply 1 (blue digit) with all digits of first number but start writing result from the position of blue digit:

$\begin{array}{l@{\,}l@{\,}l@{\,}l} & & \color{brown}{3}&\color{purple}{4} \\ \times & & \color{blue}{1}&\color{green}{2} \\ \hline & & 6 &8 \\ &\color{blue}{1 \times \color{brown}{3}}&\color{blue}{1 \times \color{purple}{4}}&\end{array}$

$\begin{array}{l@{\,}l@{\,}l@{\,}l} & & \color{brown}{3}&\color{purple}{4} \\ \times & & \color{blue}{1}&\color{green}{2} \\ \hline & & 6 &8 \\ &3&4&\end{array}$

Now you need to add these two resulting numbers. But they are shifted, so just imagine that there are zeros, on places where digit is missing:

$\begin{array}{l@{\,}l@{\,}l@{\,}l} & & \color{brown}{3}&\color{purple}{4} \\ \times & & \color{blue}{1}&\color{green}{2} \\ \hline & \color{red}{0}& 6 &8 \\ &3&4&\color{red}{0}\end{array}$

$\begin{array}{l@{\,}l@{\,}l@{\,}l} & & \color{brown}{3}&\color{purple}{4} \\ \times & & \color{blue}{1}&\color{green}{2} \\ \hline & 0& 6 &8 \\+ &3&4&0 \\ \hline & 4&0 &8\end{array}$

So, ${34}\times{12}={408}$.

Let's do a harder example.

Example 2. Multiply 852 by 697.

First, we write number one under another:

$\begin{array}{l@{\,}l@{\,}l@{\,}l} & 8 & 5 & 2 \\ \times & 6 & 9 &7 \\ \hline & & & \\\end{array}$

Multiply 7 with all digits of first number.

For this multiply 7 and 2. Result is 14. We need only 1 digit, so we take 4 and remember 1.

$\begin{array}{l@{\,}l@{\,}l@{\,}l} & 8 & \overset{\color{green}{+1}}{5} & \color{green}{2} \\ \times & 6 & 9 &\color{green}{7} \\ \hline & & & \color{green}{4}\\\end{array}$

Next, multiply 7 and 5. Result is 35. Add remembered 1. Result is 36. Again we have two digits. So take 6 and remember 3.

$\begin{array}{l@{\,}l@{\,}l@{\,}l} & \overset{\color{blue}{+3}}{8} & {\color{blue}{5}} & 2 \\ \times & 6 & 9 &\color{blue}{7} \\ \hline & & \color{blue}{6}& 4\\\end{array}$

Now, multiply 7 and 8. Result is 56. Add remembered 3. Result is 59. Again we have two digits, but we are done with last digit of second number, i.e. 7, so just write 59.

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l} & & \color{red}{8} & 5 & 2 \\ \times & & 6 & 9 &\color{red}{7} \\ \hline & \color{red}{5} & \color{red}{9}& 6& 4\\\end{array}$

Next, work with 9. Multiply 9 by 2. Result is 18. Take 8 and remember 1.

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l} & & 8 & \overset{\color{green}{+1}}{5} & \color{green}{2} \\ \times & & 6 & \color{green}{9} &7 \\ \hline & 5 & 9& 6& 4\\ & & & \color{green}{8}&\\\end{array}$

Multiply 9 by 5. Result is 45. Add remembered 1. Result is 46. Take 6 and remember 4.

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l} & & \overset{\color{blue}{+4}}{8} & \color{blue}{5} & 2 \\ \times & & 6 & \color{blue}{9} &7 \\ \hline & 5 & 9& 6& 4\\ & & \color{blue}{6}& 8&\\\end{array}$

Multiply 9 by 8. Result is 72. Add remembered 4. Result is 76. Since we are done with 9 then write 76.

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l} & & & \color{red}{8} & 5 & 2 \\ \times & & & 6 & \color{red}{9} &7 \\ \hline & & 5 & 9& 6& 4\\ & \color{red}{7}& \color{red}{6}& 6& 8&\\\end{array}$

Finally, deal with 6. Multiply 6 and 2. Result is 12. Take 2 and remember 1.

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l} & & & 8 & \overset{\color{green}{+1}}{5} & \color{green}{2} \\ \times & & & \color{green}{6} & 9 &7 \\ \hline & & 5 & 9& 6& 4\\ & 7& 6& 6& 8&\\ & & &\color{green}{2} & &\\\end{array}$

Multiply 6 by 5. Result is 30. Add remembered 1. Result is 31. Take 1 and remember 3.

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l} & & & \overset{\color{blue}{+3}}{8} & \color{blue}{5} & 2 \\ \times & & & \color{blue}{6} & 9 &7 \\ \hline & & 5 & 9& 6& 4\\ & 7& 6& 6& 8&\\ & & \color{blue}{1}&2 & &\\\end{array}$

At last multiply 6 by 8. Result is 48. Add remembered 3. Result is 51. We are done with 6, so write 51.

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l} & & & & \color{red}{8} & 5 & 2 \\ \times & & & & \color{red}{6} & 9 &7 \\ \hline & & & 5 & 9& 6& 4\\ & & 7& 6& 6& 8&\\ & \color{red}{5}&\color{red}{1} & 1&2 & &\\\end{array}$

Now, imagine zeros on places of missing digits:

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l} & & & & 8 & 5 & 2 \\ \times & & & & 6 & 9 &7 \\ \hline &\color{red}{0} &\color{red}{0} &5 & 9& 6& 4\\ &\color{red}{0} & 7& 6& 6& 8&\color{red}{0}\\ & 5&1 & 1&2 &\color{red}{0} &\color{red}{0}\\\end{array}$

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l@{\,}l} & & & & 8 & 5 & 2 \\ \times & & & & 6 & 9 &7 \\ \hline &0 &0 &5 & 9& 6& 4\\+ &0 & 7& 6& 6& 8&0\\ +& 5&1 & 1&2 &0&0\\ \hline&5&9&3&8&4&4\end{array}$

So, ${852}\times{697}={593844}$.

Next, arises question how to multiply numbers that have different number of digits? Actually the same way, just write number with smaller number of digits under the number with larger number of digits.

Example 3. Multiply 23 by 233.

Write 23 under 233 since 23 has smaller number of digits:

$\begin{array}{l@{\,}l@{\,}l@{\,}l} &2& 3&3 \\ \times & &2&3 \\ \hline & & & \\ \end{array}$

We will do this example faster:

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l} & &\color{red}{2}& \color{brown}{3}&\color{purple}{3} \\ \times & & &\color{blue}{2}&\color{green}{3} \\ \hline & & \color{green}{3 \times \color{red}{2}}& \color{green}{3 \times \color{brown}{3}}&\color{green}{3 \times \color{purple}{3}} \\ & \color{blue}{2 \times \color{red}{2}}& \color{blue}{2 \times \color{brown}{3}}&\color{blue}{2 \times \color{purple}{3}}& \\\end{array}$

Or

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l} & &2& 3&3 \\ \times & & &2&3 \\ \hline & &6& 9&9 \\ & 4& 6&6& \\\end{array}$

$\begin{array}{l@{\,}l@{\,}l@{\,}l@{\,}l} & &2& 3&3 \\ \times & & &2&3 \\ \hline & 0&6& 9&9 \\ +& 4& 6&6& 0\\ \hline &5&3&5&9 \end{array}$

So, ${23}\times{233}={5359}$ .

Finally, lat's see how to multiply more than two numbers.

It is easy, just requires more work.

For example, suppose you need to multiply 25, 31 and 54. Take 25 and 31. Multiply them. Result is 775 (verify!). Now multiply 775 and 31. Result is 24025 (verify!).

So, ${25}\times{31}\times{54}={24025}$.

It is also worth noting that order of numbers is not important. You could easily take 31 and 54, multiply them and then multiply result by 25.

Now, it is your turn. Take pen and paper and solve following problems:

Exercise 1. Find ${32}\times{31}$.

Exercise 2. Find ${58}\times{12}$.

Exercise 3. Find ${523}\times{86}$.
Exercise 4. Find ${567}\times{228}$.
Exercise 5. Find ${745}\times{51}\times{99}\times{4581}$.