# 2.5 Divisibility Rules

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:

- \(5\)
- \(4\)
- \( 8\)

**Solution:**

- 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.

8241122**84**

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.

824112**284**

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 \( 1234*x*51234 \) 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:**

Divisibility by \(13:\)

\(13 \mid x\) iff \(13\) divides the absolute difference between \(a-4b\), 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:**