\( \ \ \ \ \ \ \) We all have learned about divisibility tests for many certain numbers in our elementary school. In this note, I will be discussing about the method that those tests are derived from. For every derivation that I will be discussing, we will be taking the variable number of the form

\[\large N = \overline {a_n a_{n-1} a_{n-2} \cdots a_2 a_1 a_0}\]

\[\text{OR}\]

\[N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0\]

I will just present you with those special cases that you are familiar with and you can apply this idea to create divisibility test for any positive integer you may want to. Also I will be using the term 'number' only and only for the set of positive integers or natural numbers in this context. Also keep in mind that you must have at least some basic knowledge of modular arithmetic to understand the derivations.

Let's begin!

\[\large \underline{\text{Divisibility by 2}}\]

**Statement**

Any number with \(2, \ 4, \ 6, \ 8\) and \(0\) as the unit digit is divisible by \(2\).

**Proof**

We have,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 \)

Taking \(\pmod{2}\) of \(N\) we get,

\(\begin{align} N & \equiv 0+0+0+ \cdots +0+0+ a_0 \pmod{2} \qquad \qquad \qquad \small \color{blue}{\text{As} \ 10^k \ \text{where} \ k \ge 1 \ \text{is always divisible by} \ 2} \\ N & \equiv a_0 \pmod{2} \end{align}\)

Therefore, \(N \equiv 0 \pmod{2}\) if \(a_0 \equiv 0 \pmod{2}\).

Since, \(a_0\) is a single digit number. The only values that satisfy are \(0, \ 2, \ 4, \ 6\) and \(8\).

*The same approach can be used for \(5\) and \(10\) as well due to the fact that \(10^k\) where \(k \ge 1\) is always divisible by \(5\) and \(10\) as well and hence, the values fitting for \(a_0\) in this case are \(0\) and \(5\) for the number \(5\) and \(0\) for the number \(10\), thus proving the divisibility tests of \(5\) and \(10\).*

\[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Q.E.D.}\]

\[\large \underline{\text{Divisibility by 3}}\]

**Statement**

Any number whose sum of digits is divisible by \(3\) is also divisible by \(3\).

**Proof**

We have,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 \)

Taking \(\pmod{3}\) of \(N\) we get,

\(\begin{align} N & \equiv 1 \times a_n\pmod{3} + 1 \times a_{n-1}\pmod{3} + 1 \times a_{n-2}\pmod{3} + \cdots + 1 \times a_2\pmod{3} + 1 \times a_1\pmod{3} + 1 \times a_0\pmod{3} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \small \color{blue}{\text{As} \ 10^k-1 \ \text{where} \ k \ge 1 \ \text{is always divisible by} \ 3} \\ & \equiv \left( a_n + a_{n-1} + a_{n-2} + \cdots + a_2 + a_1 + a_0 \right) \pmod{3} \end{align}\)

Therefore, \(N \equiv 0 \pmod{3}\) if \( a_n + a_{n-1} + a_{n-2} + \cdots + a_2 + a_1 + a_0 \equiv 0 \pmod{3}\).

Thus, the sum of digits must be divisible by \(3\) for the number to be divisible by \(3\).

*The same approach can be used for \(9\) as well due to the fact that \(10^k - 1\) where \(k \ge 1\) is always divisible by \(9\) as well and hence, the sum of digits of the number in this case must be divisible by \(9\) so that the number itself is divisible by \(9\), thus proving the divisibility test of \(9\).*

\[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Q.E.D.}\]

\[\large \underline{\text{Divisibility by 4}}\]

**Statement**

Any number whose digits in the tens and units place taken in that order respectively, if divisible by \(4\), then the number is also divisible by \(4\).

Eg- The number \(11564\) is divisible by \(4\) because \(64\) is divisible by \(4\).

**Proof**

We have,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 \)

Taking \(\pmod{4}\) of \(N\) we get,

\(\begin{align} N & \equiv 0+0+0+ \cdots +0+10 a_1+ a_0 \pmod{4} \qquad \qquad \qquad \small \color{blue}{\text{As} \ 10^k \ \text{where} \ k \ge 2 \ \text{is always divisible by} \ 4} \\ N & \equiv 10 a_1 + a_0 \pmod{4} \end{align}\)

Therefore, \(N \equiv 0 \pmod{4}\) if \(10a_1 + a_0 = \overline {a_1 a_0} \equiv 0 \pmod{4}\).

Thus, the if the tens and units place of a number taken in that order is divisible by \(4\) then the number is also divisible by \(4\).

*The same approach can be used for \(25\) as well due to the fact that \(10^k\) where \(k \ge 2\) is always divisible by \(25\) as well and hence, if the digits in the tens and units place of a number taken in that order respectively, if divisible by \(25\), then the number is also divisible by \(25\).*

\[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Q.E.D.}\]

\[\large \underline{\text{Divisibility by 6}}\]

**Statement**

Any number which is divisible by \(2\) and \(3\) both is also divisible by \(6\) as well.

**Proof**

This doesn't require any proof other that the fact that if

\(N \equiv 0 \pmod{2}\) and \(N \equiv 0 \pmod{3}\), then \(N \equiv 0 \pmod{2 \times 3 = 6}\). As \(2\) and \(3\) are co-prime numbers.

\[\large \underline{\text{Divisibility by 7}}\]

**Statement**

Any number whose absolute difference between twice the units digit and the number formed by the rest of the digits, if it is \(0\) or divisible by \(7\), then the number is also divisible by \(7\).

Eg- The number \(343\) is divisible by \(7\) because \(34 - 2 \times 3 = 28\) is also divisible by \(7\).

**Proof**

We have,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 \)

Let,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 = 7k \qquad \quad \small \color{blue}{\text{for some integer} \ k}\)

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 \right) + a_0 = 7k\)

Add and subtract \(20 a_0\) on the L.H.S. to get,

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 \right) + 20 a_0 - 20 a_0 + a_0 = 7k\)

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 - 2 a_0 \right) + 21 a_0 = 7k\)

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 - 2 a_0 \right) = 7k - 21 a_0 \)

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 - 2 a_0 \right) \equiv 0 \pmod{7}\)

\(N = 10 \left( \overline{a_n a_{n-1} a_{n-2} \cdots a_2 a_1} - 2 a_0 \right) \equiv 0 \pmod{7}\)

Since, \(10 \equiv 3 \pmod{7}\), therefore, for \(N\) to be divisible by \(7\), \(\overline{a_n a_{n-1} a_{n-2} \cdots a_2 a_1} - 2 a_0 \equiv 0 \pmod{7}\).

Thus, for a number if the absolute difference between twice the units digit and the number formed by the rest of the digits, is \(0\) or divisible by \(7\), then that number is also divisible by \(7\).

\[\large \underline{\text{Divisibility by 8}}\]

**Statement**

Any number whose digits in the hundreds, tens and units place taken in that order respectively, if divisible by \(8\), then the number is also divisible by \(8\).

Eg- The number \(74152\) is divisible by \(8\) because \(152\) is divisible by \(8\).

**Proof**

We have,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 \)

Taking \(\pmod{8}\) of \(N\) we get,

\(\begin{align} N & \equiv 0+0+0+ \cdots +10^2 a_2+10 a_1+ a_0 \pmod{8} \qquad \qquad \qquad \small \color{blue}{\text{As} \ 10^k \ \text{where} \ k \ge 3 \ \text{is always divisible by} \ 8} \\ N & \equiv 100 a_2 + 10 a_1 + a_0 \pmod{8} \end{align}\)

Therefore, \(N \equiv 0 \pmod{8}\) if \(100 a_2 + 10a_1 + a_0 = \overline {a_2 a_1 a_0} \equiv 0 \pmod{8}\).

Thus, the if the hundreds, tens and units place of a number taken in that order is divisible by \(8\) then the number is also divisible by \(8\).

*The same approach can be used for \(125\) as well due to the fact that \(10^k\) where \(k \ge 3\) is always divisible by \(125\) as well and hence, if the digits in the hundreds, tens and units place of a number taken in that order respectively, if divisible by \(125\), then the number is also divisible by \(125\).*

\[\large \underline{\text{Divisibility by 11}}\]

**Statement**

Any number whose absolute difference of the sum of digits occurring in the even positions and the sum of digits occurring in odd positions, if it is \(0\) or divisible by \(11\), then the number is also divisible by \(11\).

**Proof**

We have,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 \)

Taking \(\pmod{11}\) of \(N\) we get,

\(N \equiv \pm a_n \mp a_{n-1} \pm a_{n-2} \cdots \mp a_2 \pm a_1 \mp a_0 \pmod{11} \qquad \qquad \small \color{blue}{\text{As} \ 10^k \equiv +1 \pmod{11} \ \text{if} \ k \ \text{is even and} \ 10^k \equiv -1 \pmod{11} \ \text{if} \ k \ \text{is odd}}\)

- Suppose \(n\) is even, then we have:

\(N \equiv a_n - a_{n-1} + a_{n-2} - \cdots + a_2 - a_1 + a_0 \pmod{11} \\ N \equiv \left( a_n + a_{n-2} + \cdots + a_2 + a_0 \right) - \left( a_{n-1} + a_{n-3} + \cdots + a_3 + a_1 \right) \pmod{11}\)

Therefore, \(N \equiv 0 \pmod{11}\) if \(\left( a_n + a_{n-2} + \cdots + a_2 + a_0 \right) - \left( a_{n-1} + a_{n-3} + \cdots + a_3 + a_1 \right) \equiv 0 \pmod{11}\), given that \(n\) is even.

- Suppose \(n\) is odd, then we have:

\(N \equiv -a_n + a_{n-1} - a_{n-2} + \cdots + a_2 - a_1 + a_0 \pmod{11} \\ N \equiv \left( a_{n-1} + a_{n-3} + \cdots + a_2 + a_0 \right) - \left( a_n + a_{n-2} + \cdots + a_3 + a_1 \right) \pmod{11}\)

Therefore, \(N \equiv 0 \pmod{11}\) if \(\left( a_{n-1} + a_{n-3} + \cdots + a_2 + a_0 \right) - \left( a_n + a_{n-2} + \cdots + a_3 + a_1 \right) \equiv 0 \pmod{11}\), given that \(n\) is odd.

From the above two conditions, we infer that for a number to be divisible by \(11\), its absolute difference of the sum of digits occurring in the even positions and the sum of digits occurring in odd positions, should be \(0\) or divisible by \(11\).

\[\large \underline{\text{Divisibility by 12}}\]

**Statement**

Any number which is divisible by \(3\) and \(4\) both is also divisible by \(12\) as well.

**Proof**

This doesn't require any proof other that the fact that if

\(N \equiv 0 \pmod{3}\) and \(N \equiv 0 \pmod{4}\), then \(N \equiv 0 \pmod{3 \times 4 = 12}\). As \(3\) and \(4\) are co-prime numbers.

\[\large \underline{\text{Divisibility by 13}}\]

**Statement**

Any number whose sum of four times the units digit and the number formed by the rest of the digits, if it is divisible by \(13\), then the number is also divisible by \(13\).

Eg- The number \(169\) is divisible by \(13\) because \(16 + 4 \times 9 = 52\) is also divisible by \(13\).

**Proof**

We have,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 \)

Let,

\(N = 10^n a_n + 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^2 a_2 + 10 a_1 + a_0 = 13k \qquad \quad \small \color{blue}{\text{for some integer} \ k}\)

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 \right) + a_0 = 13k\)

Add and subtract \(40 a_0\) on the L.H.S. to get,

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 \right) + 40 a_0 - 40 a_0 + a_0 = 13k\)

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 + 4 a_0 \right) - 39 a_0 = 13k\)

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 + 4 a_0 \right) = 13k + 39 a_0 \)

\(N = 10 \left( 10^{n-1} a_n + 10^{n-2} a_{n-1} + 10^{n-3} a_{n-2} + \cdots + 10 a_2 + a_1 + 4 a_0 \right) \equiv 0 \pmod{13}\)

\(N = 10 \left( \overline{a_n a_{n-1} a_{n-2} \cdots a_2 a_1} + 4 a_0 \right) \equiv 0 \pmod{13}\)

Since, \(10 \equiv 10 \pmod{13}\), therefore, for \(N\) to be divisible by \(13\), \(\overline{a_n a_{n-1} a_{n-2} \cdots a_2 a_1} + 4 a_0 \equiv 0 \pmod{13}\).

Thus, for a number if the sum of four times the units digit and the number formed by the rest of the digits, is divisible by \(13\), then that number is also divisible by \(13\).

Based on the same approach, a divisibility test can be made for each and every number by just observing their patterns over successive powers of \(10\).

Hope you enjoyed reading this.

\[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Thank \ you\]

## Comments

Sort by:

TopNewestWOOOOOAHHH! You just shown the proof for most of the divisibility rules. You should post your note as another separate wiki called "Proof of divisibilities rules"! – Pi Han Goh · 7 months, 3 weeks ago

Log in to reply

here. Feel free to share your opinion about the wiki and it would be very kind of you if you can contribute by citing problems and examples. – Tapas Mazumdar · 7 months, 3 weeks ago

Thanks for encouraging me to write this wiki. You can check it outLog in to reply

Btw, thanks. :) – Tapas Mazumdar · 7 months, 3 weeks ago

Log in to reply

Proof Of Divisibility Rules and we can link it from there. – Calvin Lin Staff · 7 months, 3 weeks ago

I agree that adding the proofs for these divisibility rules will be helpful for those who want to learn about the underlying number theoretic. It would be great if you can contribute toLog in to reply

here. Feel free to share your opinion about the wiki and it would be very kind of you if you can contribute by citing problems and examples. – Tapas Mazumdar · 7 months, 3 weeks ago

Thanks for encouraging me to write this wiki. You can check it outLog in to reply

Excellent!

I hope my grade for 4 and 5 teachers knew it, I always troubled them asking for the proof :P – Swapnil Das · 7 months, 2 weeks ago

Log in to reply

– Tapas Mazumdar · 7 months, 2 weeks ago

Haha, well I had asked my teachers too about this once. But they told that it wasn't in our syllabus and shut my mouth.Log in to reply

– Harsh Shrivastava · 7 months, 2 weeks ago

Syllabus based school studies is what preventing India from becoming a developing country to a developed country.Log in to reply

– Tapas Mazumdar · 7 months, 2 weeks ago

That is a weak part of our system. If a student is keen to learn more about a certain concept, their teachers must put their best efforts to help them go through these concepts. Moreover, it is the fact that our teachers have the burden of keeping things straight to syllabus and that's part of the problem that even they don't think it is important for them to help a child go through the concepts that are beyond the school syllabus.Log in to reply

Nicely written! – Harsh Shrivastava · 7 months, 3 weeks ago

Log in to reply

– Tapas Mazumdar · 7 months, 3 weeks ago

Thank you! :DLog in to reply