maths > wholedivisors

Simple Divisibility Tests

what you'll learn...

overview

•  the divisibility test for 2$2$

•  the divisibility test for 10$10$

•  the divisibility test for 3$3$

•  the divisibility test for 4$4$

•  the divisibility test for 5$5$

•  the divisibility test for 11$11$

•  the divisibility test for $9$

•  the divisibility test for $6$

Divisibility by 2

Consider the multiples of $2$ given as $2 , 4 , 6 , 8 , 10 , 12 , \cdots$. A similarity is observed in the multiples -- they are all even numbers.

Let us consider the numbers $4$ and $5$.
The even number $4$ is divisible by $2$. But the odd number $5$ is not divisible by$2$.

Even numbers are divisible by $2$. To check for divisibility by $2$, only the digit in the units place is checked.

Test for Divisibility by $2$ : If the digit in the units place is even, then the number is divisible by $2$.

Is $218$ divisible by $2$?
"Yes". Checking the digit in units place $8$, it is concluded that the number is even and so divisible by $2$.

Is $23$ divisible by $2$?
"No". Checking the digit in units place $3$, it is concluded that the number is odd and so NOT divisible by $2$.

Divisibility by 10

Consider the multiples of $10$ given as $10 , 20 , 30 , 40 , \cdots$. A similarity is observed in the multiples -- "they all have digit $0$ in the units place". That is, all multiples of $10$ end in $0$.

Numbers with $0$ in units place are divisible by $10$.

Consider the numbers $20$ and $24$. Checking the units place, $20$ has $0$ and so $20$ is divisible by $10$. But $24$ has $4$ in units place, and so $24$ is not divisible by 10.

Test for Divisibility by $10$ : If the digit in the units place is $0$, then the number is divisible by $10$.

Is $2008$ divisible by $10$?
"No". Checking the digit in units place $8$, it is concluded that the number is not divisible by $10$.

Is $920$ divisible by $10$?
"Yes". Checking the digit in units place $0$, it is concluded that the number is divisible by $10$.

Divisibility by 3

"$3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , \cdots$" are the multiples of $3$. From the list, it is not easy to find a common property among the multiples.

Consider the multiples of $3$ given as $3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , \cdots$. The similarity is "sum of all digits is one of the numbers $3 , 6 , 9$".

This is explained as follows
$12 \to 1 + 2 = 3$ sum is $3$
$15 \to 1 + 5 = 6$ sum is $6$
$18 \to 1 + 8 = 9$ sum is $9$
$21 \to 2 + 1 = 3$ sum is $3$
$24 \to 2 + 4 = 6$ sum is $6$
$27 \to 2 + 7 = 9$ sum is $9$

The property is true for all multiples of $3$.

To identify the multiplies of $3$, the sum of all digits of the number is checked for divisibility by $3$.

Explanation for the curious mind.

Consider divisibility test $42$ by $3$

To simplify the divisibility test, let us subtract a multiple of the divisor $3$.
Seeing the $10$s place value $4$, we choose the multiple $3 \times 3 \times 4 = 36$ to subtract from the number.

As per the property of simplification by subtraction, the divisibility test of $42$ is simplified into divisibility test of $42 - 36 = 40 - 36 + 2 = 4 + 2$.

This explains the divisibility test for $3$.

Consider the numbers $12$ and $13$.
$12 \to 1 + 2 = 3$ sum $3$ is divisible by 3, so $12$ is divisible by $3.$
$13 \to 1 + 3 = 4$ sum $4$ is not divisible by 3, so $13$ is not divisible by $3.$

Test for Divisibility by 3 : If the sum of all the digits is divisible by $3$ then the number is divisible by $3$

Is $318$ divisible by $3$?
Yes. Checking the sum of digits $3 + 1 + 8 = 12$. The sum is divisible by $3$. It is concluded that the number is divisible by $3$.

Is $923$ divisible by $3$?
No. Checking sum of the digits $9 + 2 + 3 = 14$, it is concluded that the number is NOT divisible by $3$.

Divisibility by 4

The multiples of $4$ have the property that last two digits are divisible by $4$.

Explanation for the curious mind

A large number like $2344$ can be given as $23 \times 100 + 44$.

It was learned that if one addend is divisible, then the divisibility is decided by the other addend.

$23 \times 100$ is always divisible by $4$ as $100 = 25 \times 4$.

So, the divisibility test is done in $44$, that is the tens and units digits.

Test for Divisibility by 4 : If the last $2$ digits (tens and units place value positions) of the number is divisible by $4$ then the number is divisible by $4$.

Is $2018$ divisible by $4$?
No. Checking the tens and units digits $18$, it is concluded that the number is not divisible by $4$.

Is $920$ divisible by $4$?
Yes. Checking the tens and units digits $20$, it is concluded that the number is divisible by $4$.

Divisibility by 5

the multiples of $5$ are $5 , 10 , 15 , 20 , 25 , \cdots$. The multiples of $5$ has either $5$ or $0$ in the units place. The divisibility test for $5$ is to check if the units digit is either $5$ or $0$.

Explanation for the curious mind

A large number like $2344$ can be given as $234 \times 10 + 4$.

It was learned that if one addend is divisible, then the divisibility is decided by the other addend.

$234 \times 10$ is divisible by $5$ as $10 = 5 \times 2$.

So, the divisibility test is done on the last digit, that is the units digits. In the units digit the possible values are $5$ and $0$.

Consider $10$ and $25$. Both $10$ and $25$ are divisible by $5$
$10$ has $0$ in units place. So, $10$ is divisible by $5$.
$25$ has $5$ in units place. So, $25$ is divisible by $5$.

Test for Divisibility by $5$ : If the units digit is $5$ or $0$, then the number is divisible by $5$.

Is $2018$ divisible by $5$?
No. Checking the digit in units place $8$, it is concluded that the number is not divisible by $5$.

Is $920$ divisible by $5$?
Yes. Checking the digit in units place $0$, it is concluded that the number is divisible by $5$.

Divisibility by 11

Consider the multiples of $11$ given as $11 , 22 , 33 , \cdots , 110 , 121 , 132 , \cdots$. Is there any similarity observed in the multiples?
The answer is "alternate digits have some pattern".

"Note: A number $3521$ when multiplied by $11$ is given as

$3521 \times 11$
$= 3521 \times \left(10 + 1\right)$
$= 35210 + 3521$
$= \left(3\right) \left(5 + 3\right) \left(2 + 5\right) \left(1 + 2\right) \left(1\right)$

Four digit number, when multiplied by $11$, results in
• units digit of product is same as that of multiplicand
• tens digit of product is sum of units and tens digit of multiplicand
• hundreds digit of product is sum of tens and hundreds digit of multiplicand
• and so on.
We can use this property to device a divisibility test for $11$.

The multiples of $11$ have the property explained below. For example, consider $121$ (a multiple of $11$).

Add the digits in odd positions $1 + 1$, the result is $2$
Add the digits in even positions $2$.
Find the difference between these two results.
If the difference is $0$ or a multiple of $11$, then the number is a divisible by $11$.

Test for Divisibility by 11 : Find the sum of digits in even positions and the sum of digits in odd positions. If the difference between these two sums is a multiple of 11, then the number is divisible by 11.

Is $2108$ divisible by $11$?
No. The sum of alternate digits are $2 + 0 = 2$ and $1 + 8 = 9$. The difference between these sums are $9 - 2 = 7$. Since $7$ is not divisible by $11$, the number is not divisible by $11$.

Is $902$ divisible by $11$?
Yes. The $9 + 2 - 0 = 11$, so the number is divisible by $11$.

Divisibility by 9

Consider the multiples of $9$ given as $9 , 18 , 27 , 36 , \cdots$. A similarity is observed in these multiples : "sum of all digits is a multiple of $9$".

This is explained as follows
$18 \to 1 + 8 = 9$ sum is $9$
$27 \to 2 + 7 = 9$ sum is $9$
$36 \to 3 + 6 = 9$ sum is $9$
$189 \to 1 + 8 + 9 = 18$ sum is multiple of $9$

This is true for all multiples of $9$.

To identify the multiplies of $9$, the sum of all digits of the number is checked for divisibility by $9$.

Explanation for the curious mind.

Consider divisibility test $42$ by $9$

To simplify the divisibility test, let us subtract a multiple of the divisor $9$.
Seeing the $10$s place value $4$, we choose the multiple $9 \times 4$ to subtract from the number.

As per the property of simplification by subtraction, the divisibility test of $42$ is simplified into divisibility test of $42 - 36 = 40 - 36 + 2 = 4 + 2$.

This explains the divisibility test for $9$.

Test for Divisibility by $9$ : If the sum of all the digits is divisible by $9$, then the number is divisible by $9$

Is $2008$ divisible by $9$?
No. Checking the sum of digits $2 + 0 + 0 + 8 = 10$, it is concluded that the number is not divisible by $9$.

Is $927$ divisible by $9$?
Yes. Checking the sum of digits $9 + 2 + 7 = 18$, it is concluded that the number is divisible by $9$.

Divisibility by 6

Consider the number $3 \times 2 \times 2351$, is it divisible by $6$?
Yes, it is divisible by $6$ as it has $3$ and $2$ as factors.

Test for Divisibility by $6$ : If the number is divisible by $2$ and divisible by $3$ then the number is divisible by $6$

Is $2008$ divisible by $6$?
No. The number is divisible by $2$ but not by $3$. So it is not divisible by $6$.

Is $912$ divisible by $6$?
Yes. The number is divisible by $2$ and also divisible by $3$. So it is divisible by $6$.

summary

Test for Divisibility by $2$ : If the digit in the units place is even, then the number is divisible by $2$.

Test for Divisibility by $10$ : If the digit in the units place is $0$, then the number is divisible by $10$.

Test for Divisibility by 3 : If the sum of all the digits is divisible by $3$ then the number is divisible by $3$

Test for Divisibility by 4 : If the last $2$ digits (tens and units place value positions) of the number is divisible by $4$ then the number is divisible by $4$.

Test for Divisibility by $5$ : If the units digit is $5$ or $0$, then the number is divisible by $5$.

Test for Divisibility by 11 : Find the sum of digits in even positions and the sum of digits in odd positions. If the difference between these two sums is a multiple of 11, then the number is divisible by 11.

Test for Divisibility by $9$ : If the sum of all the digits is divisible by $9$, then the number is divisible by $9$

Test for Divisibility by $6$ : If the number is divisible by $2$ and divisible by $3$ then the number is divisible by $6$

Outline

The outline of material to learn "Divisibility in Whole Numbers" is as follows.

Note: click here for detailed outline of Whole divisors

→   Classification as odd, even, prime, and composite

→   Factors, Multiples, Prime factorization

→   Highest Common Factor

→   Lowest Common Multiple

→   Introduction to divisibility tests

→   Simple Divisibility Tests

→   Simplification of Divisibility Tests

→   Simplification in Digits for Divisibility Tests