
# 2.5 Divisibility Rules

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In this section, we will explore the rules of divisibility for 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 $$a= d_n10^{n-1} +d_{n-1}10^{n-2}+ \cdots+ d_2 10+d_1$$.

Divisibility by $$2=2^1:$$

$$2 \mid x$$ 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 $$8=2^3:$$

$$8 \mid x$$ 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, it’s 4, then 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.

824112284

à (4)(21) = 84

Since 84 is divisible by 4, then the original number, 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.

824112284

à (8)(35) = 280

Since 284 is not divisible by 8, then the original number, 824112284 is not divisible by 8 either.

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

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

Example $$\PageIndex{2}$$:

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

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.

$$2(1 + 2 + 3 + 4) + 5$$

$$= 2(10) + 5$$

$$= 20 + 5$$

$$= 25$$

The number $$25$$ is not divisible by 3, but 27, 30, and 33 are.

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:

Divisibility by $$11:$$

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

Proof: