# 2.5: Divisibility Rules

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In this section, we will explore some rules of divisibility for some positive integers.  These rules can be easily extended to all the integers by dropping the sign.

Let $$x \in \mathbb{Z_+}$$.

Then $x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0,$

which implies

$x= 10( d_n10^{n-1} +d_{n-1}10^{n-2}+ \cdots+ d_2 10+d_1)+d_0.$

Thus we can express $$x$$ as $$10 a+b$$, where $$b=d_0$$ is the ones digit of $$x$$, and $$a= d_n10^{n-1} +d_{n-1}10^{n-2}+ \cdots+ d_2 10+d_1$$.

##### Divisibilty by $$2=2^1$$:

Let $$x \in \mathbb{Z_+}$$. $$2 \mid x$$ iff $$2 \mid b$$. In other words, $$2$$ divides an integer iff the ones digit of the integer is either $$0, 2, 4, 6,$$ or $$8$$.

Proof:

Since $$2\mid 10$$, $$x=10 a+b$$, and by divisibility theorem I, $$2 \mid x$$ iff $$2 \mid b$$.$$\Box$$

##### Divisibility by $$5:$$

$$5 \mid x$$ iff $$5 \mid b$$. In other words, $$5$$ divides an integer iff the ones digit of the integer is either $$0,$$ or $$5$$.

Proof:

Since $$5 \mid 10$$, $$x=10 a+b$$, and by divisibility theorem I, $$5 \mid x$$ iff $$5 \mid b$$.$$\Box$$

##### Divisibility by $$10:$$

$$10 \mid x$$ iff $$10 \mid b$$. In other words, $$10$$ divides an integer iff the ones digit of the integer is $$0,$$.

Proof:

Since $$10 \mid 10$$, $$x=10 a+b$$, and by divisibility theorem I, $$10 \mid x$$ iff $$10 \mid b$$.$$\Box$$

##### Divisibility by $$4=2^2:$$

$$4 \mid x$$ iff $$4 \mid d_1d_0$$.

Proof:
Let $$x$$ be an integer. Then

$$x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0$$, which implies

$$x= 100( d_n10^{n-2} +d_{n-1}10^{n-3}+ \cdots+ d_2)+ 10d_1+d_0=100( d_n10^{n-2} +d_{n-1}10^{n-3}+ \cdots+ d_2)+ d_1d_0$$.

Since $$4 \mid 100$$, $$x=100 a+ d_1d_0$$, and by divisibility theorem I, $$4 \mid x$$ iff $$4 \mid d_1d_0$$.$$\Box$$

##### Divisibility by $$4$$ of a two digit number

$$4 \mid d_1d_0$$ iff $$4 \mid 2d_1+d_0$$

##### Divisibility by $$8=2^3:$$

$$8 \mid x$$ iff $$8 \mid d_2 d_1d_0$$.

Proof:

Let $$x$$ be an integer. Then

$$x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0$$, which implies

$$x= 10^3( d_n10^{n-3} +d_{n-1}10^{n-4}+ \cdots+ d_3)+ 100d_2+10d_1+d_0=10^3( d_n10^{n-3} +d_{n-1}10^{n-4}+ \cdots+ d_3)+ d_2d_1d_0$$.

Since $$8 \mid 1000$$, $$x=1000 a+ d_2d_1d_0$$, and by divisibility theorem I, $$8 \mid x$$ iff $$8 \mid d_2d_1d_0$$.$$\Box$$

A similar argument can be made for divisibility by $$2^n$$, for any positive integer $$n$$.

Example $$\PageIndex{1}$$:

Using divisibility tests, check if the number $$824112284$$ is divisible by:

1. $$5$$
2. $$4$$
3. $$8$$

Solution:

1.   $$824112284$$ is not divisible by $$5.$$

Rule: The one's digit of the number has to be either a $$0$$ or a $$5.$$

Since the last digit is not $$0$$ or $$5,$$  $$824112284$$  is not divisible by $$5.$$

2. $$824112284$$ is divisible by $$4.$$

Rule: The last two digits of the number have to be divisible by 4.

The last two digits of $$824112284$$ is $$84=(4)(21).$$

Since $$84$$ is divisible by $$4, 824112284$$ is divisible by $$4$$ also.

3. $$824112284$$ is not divisible by $$8.$$

Rule: The last three digits of the number have to be divisible by 8.

The last three digits of $$824112284$$ is $$284.$$ Since $$284$$ is not divisible by $$8,$$  the original number, $$824112284$$ is not divisible by $$8$$ either.

Notice the following pattern $$10-1=9, 100-1=99, 1000-1=999, \cdots, 10^n-1=999 \cdots 9$$

##### Divisibility by $$3=3^1:$$

$$3 \mid x$$ iff $$3$$ divides sum of its digits.

Proof:
Let $$x \in \mathbb{Z_+}$$.

Then $x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0,$

which implies

$x= 9( d_n(10^{n-1} +d_{n-1}10^{n-2}+ \cdots+ d_2 10+d_1)+ (d_0.$

Example $$\PageIndex{2}$$:

Find the possible values for the missing digit $$x$$, if $$1234x51234$$ is divisible by $$3.$$

Solution:

Consider the following:

The divisibility rule for the number 3 is as follows: If the sum of the digits in the whole number is a number divisible by 3, then the larger, original number is also.

The sum of the digits is $$2(1+2+3+4)+5 +x=25 +x.$$ Since $$3\mid (25+x)$$, $$25+x=27, 30$$ or $$33$$. Hence $$x=2,5$$ or $$8$$.

##### Divisibility by $$9=3^2:$$

$$9 \mid x$$ iff $$9$$ divides sum of its digits.

##### Divisibility by $$7:$$

$$7 \mid x$$ iff $$7$$ divides the absolute difference between $$a-2b$$, where $$x=10 a+b$$, where $$b=d_0$$ is the ones digit of $$x$$, and $$a= d_n10^{n-1} +d_{n-1}10^{n-2}+ \cdots+ d_2 10+d_1$$.

Proof:

Assume $$7\mid x.$$ Which implies $$7\mid 10a+b.$$

Consider $$x-21b=(10a+b)-21b=10a-20b=10(a-2b)$$.

Thus $$7\mid x$$ if and only if  $$7\mid a-2b.$$

##### Divisibility by $$11:$$

$$11 \mid x$$ iff $$11$$ divides the absolute difference between the alternate sum.

Proof:

This page titled 2.5: Divisibility Rules is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Pamini Thangarajah.