# 2.5 Divisibility Rules

- Page ID
- 7463

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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:

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

**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:**