# Multiplication table

The multiplication table includes multiplication results for numbers from $1$ to $9$.

The multiplication is a lot easier if you know the **multiplication table**.

The result of a multiplication of two numbers is called a** product**.

The numbers we multiply are called **the factors**.

The first factor is called the **multiplicand** and the second is called **multiplier**.

The multiplication of one-digit numbers is the easiest form of multiplication and must be learned using the *multiplication table* before moving on to more complex examples.

Now, we are going to look at multiplication of one-digit numbers with two-digit numbers.

Two-digit numbers are composed of numerals *ones* and *tens*.

For example, number $45$ we can write down like:

$\ 45 = 4 \cdot 10 + 5 \cdot 1$

## Multiplication of natural numbers

We are going to start with an example of multiplication of number $8$ with $53$:

We started by aligning the multiplier, number $8$, with the number $3$. Then we multiplied each digit of the multiplicand.

We wrote the “ones” digit of each product in the result. If the number has “tens” digit then it will be *carried* and added to the next product.

The process should look like this:

-> $3 \cdot 8 =24$. Write number $4$ and carry the number $2$.

-> $5 \cdot 8 = 40$. We adding carried number $2$ to number $40$. The result is number $42$.

-> The final result is number $424$.

Now, we will move on to the __multiplication of two-digit numbers__.

Let’s take a look at the next example. We will multiply number $71$ with number $56$:

First we multiplied the “ones” digit, number $6$, with number $71$ using the method shown before. The result is number $426$.

Now we multiply number $5$ with number $71$. The result is number $355$. Note that we shifted the result (355) one digit to the left, and a number $0$ is added at the end. The result is shifted because we multiplied the “tens” digit, number $5$, with $71$.

The process should look like this:

-> $6 \cdot 1= 6$. Write number $6$.

-> $6 \cdot 7 = 42$. Write number $42$. Now we get number $426$.

-> $5 \cdot 1 = 5$. Write number $5$.

-> $5 \cdot 7 =35$. Write number $35$. Now we got the number $355$, and it is shifted by one digit to the left.

That means that *the last digit is $0$*.

-> Now we need to add number $426$ to $3550$ and the result is number $3976$.