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Math Basics

Check Divisibility Rules for Numbers from 1 to 20

In this article, we'll take a closer look at divisibility rules.
These are simple tricks that help you quickly determine if one number can be divided by another, without actually doing the division.

These rules are really helpful for mental math, allowing you to solve problems faster, whether you're working with factors, simplifying fractions, or using modular arithmetic.

Each divisibility rule is based on patterns found in the digits of a number.
This means you don't need to perform long division to check if a number is divisible by 2, 3, 5, or other numbers.

Divisibility by 1

Rule: Every number is divisible by 1.
This is the simplest rule. Since dividing any number by 1 results in the number itself, all numbers satisfy this rule.

Divisibility by 2

Rule: A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.
For example, 34 ends in 4, which is even, so 34 is divisible by 2.
This is because even numbers end in these digits.

Divisibility by 3

Rule: A number is divisible by 3 if the sum of its digits is divisible by 3.
For example, To check if 123 is divisible by 3, add the digits: 1 + 2 + 3 = 6. Since 6 is divisible by 3, So 123 is too divisible by 3.

Let's understand the reasoning behind this trick.

Any decimal number N can be represented as : 

N = a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d​, where a, b, c, d are the digits of the numbers.

So if we observe a property, i.e. when you divide the power of 10 by 3 we get 1 as the remainder.

i.e. 10%3 = 1, 100%3 = 1, 1000%3 = 1 and so on.

So for any integer n, (10^n) % 3 = 1.

Now so, N = a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d, if we take the modulo of N with 3 we get,

N%3 = (a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d)%3.

= ((a​*10^k)%3 + (b*10^(k−1))%3 + ⋯ + (c​*10^1)%3 + d%3)

As for any integer n, 10^n % 3 = 1.

So our N simplified to

= (a%3 + b%3 + ⋯ + c%3 + d%3)

= (a + b + c + d)%3

i.e. Sum of digits of n should be divisible by 3.

Divisibility by 4

Rule: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

For example, for the number 132, consider the last two digits (32). Since 32 is divisible by 4, 132 is divisible by 4.

Let's understand the reasoning behind this trick.

Let's consider a decimal number N. We can express N in terms of its last two digits and the remaining part of the number:

N = 100*Q + R

where Q is the integer part of the number when the last two digits are removed, and R is the number formed by the last two digits.

To understand why only the last two digits matter, let's look at the properties of powers of 10 modulo 4:

10^1 % 4 = 2

10^2 % 4 = 0

10^3 % 4 = 0

10^4 % 4 = 0 and so on.

So for any 10^n % 4 = 0 for n>=2, this tells us that any power of 10 higher than or equal to 10^2 (i.e., 100) is divisible by 4.

When we take modulo by N, we get 

N%4 = (100*Q + R)%4 
Since 100%4 = 0 , 100*Q % 4 = 0, 
Therefore N%4 = (0 + R)%4 = R%4.

This shows that the remainder when N is divided by 4 is the same as the remainder when the last two digits R are divided by 4.

Divisibility by 5

Rule: A number is divisible by 5 if its last digit is 0 or 5.

For example, 245 ends in 5, so it is divisible by 5. Similarly, 450 ends in 0, so it is also divisible by 5.

Since the odd multiple of 5 has the last digit as 5 in it.

For example: 5*1 = 5, 5*3 = 15, 5*5 = 25, 5*7 = 35 and so on.

Even a multiple of 5 has 0 as the last digit. 

For example: 5*2 = 10, 5*4 = 20, 5*6 = 30, 5*8 = 40 and so on.

So to check whether a number is divisible by 5 or not we check its last digit whether it is 0 or 5 or not.

Divisibility by 6

Rule: A number is divisible by 6 if it is divisible by both 2 and 3.

The number 6 can be factored into prime numbers as 6=2×3.

Therefore, a number divisible by 6 must be divisible by both 2 and 3, because 6 is the product of these two factors.

For example, 54 is divisible by 2 (the last digit is 4) and 3 (54 is divisible by 3 because 5 + 4 = 9, and 9 is divisible by 3). Therefore, 54 is divisible by 6.

Divisibility by 7

Rule: Subtract twice the last digit from the rest of the number. If the result is divisible by 7 (or is 0), then the original number is divisible by 7.

For example, for 203, subtract 2 times the last digit (3) from the rest (20) : 

20 - 2*3 = 20 - 6 = 14, which is divisible by 7. 

Therefore, 203 is divisible by 7.

Let's understand the reasoning behind this trick.

  • Representation of the Number:
    • Any number n can be expressed in the form: n=10a+b
      where:
      • a is the number formed by the digits of n, excluding the last digit.
      • b is the last digit of n.
  • Modulo 7 Properties:
    • To test if n is divisible by 7, we check if n % 7=0.
  • Breaking n into Parts:
    • Substitute n=10a+b into the expression n mod 7:
      n % 7=(10a+b) % 7
    • Using the distributive property of modular arithmetic:
      n % 7=(10a % 7+b % 7) % 7
  • Simplify 10%7:
    • Since 10 % 7=3, we replace 10a % 7 with 3a%7, so n%7 = (3a+b)%7
  • Reformulate the Rule:
    • Rearrange the terms to isolate a: 3a+b≡0 (mod 7)
    • Subtract 2b (because 2b % 7=−b % 7): a−2b≡0(mod7)
    • This simplifies to: a−2b % 7=0

Interpret the Rule:
This shows that subtracting twice the last digit b from the rest of the number a preserves divisibility by 7.

If the result (a−2b) is divisible by 7 (or equals 0), then the original number n is divisible by 7.

Divisibility by 8

Rule: A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

For example, for the number 1,216, look at the last three digits (216). Since 216 is divisible by 8 (216 ÷ 8 = 27), 1,216 is divisible by 8.

Let's understand the reasoning behind this trick.

Let's consider a decimal number N. We can express N in terms of its last two digits and the remaining part of the number:

N = 1000*Q + R

where Q is the integer part of the number when the last three digits are removed, and R is the number formed by the last three digits.

To understand why only the last two digits matter, let's look at the properties of powers of 10 modulo 8:

10^1 % 8 = 2

10^2 % 8 = 4

10^3 % 8 = 0

10^4 % 8 = 0 

10^5 % 8 = 0 and so on.

So for any 10^n % 8 = 0 for n>=3, this tells us that any power of 10 higher than or equal to 10^3 

(i.e., 1000) is divisible by 8.

When we take modulo by N, we get 

N%8 = (1000*Q + R)%8 

Since 1000%8 = 0 , 1000*Q % 8 = 0, 

Therefore N%8 = (0 + R)%8 = R%8.

Divisibility by 9

Rule: A number is divisible by 9 if the sum of its digits is divisible by 9.

For example, for 729, add the digits: 7 + 2 + 9 = 18. Since 18 is divisible by 9, 729 is divisible by 9.

Let's understand the reasoning behind this trick.

Any decimal number N can be represented as : 

N = a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d​, where a, b, c, d are the digits of the numbers.

So if we observe a property, i.e. when you divide the power of 10 by 9 we get 1 as the remainder.

i.e 10%9 = 1, 100%9 = 1, 1000%9 = 1 and so on.

So for any integer n, 10^n % 9 = 1.

Now so, N = a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d, if we take the modulo of N with 9 we get,

N%9 = (a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d)%9.

= ((a​*10^k)%9 + (b*10^(k−1))%9 + ⋯ + (c​*10^1)%9 + d%9)

As for any integer n, 10^n % 9 = 1.

So our N simplified to

= (a%9 + b%9 + ⋯ + c%9 + d%9)

= (a + b + c + d)%9

i.e. Sum of digits of n should be divisible by 9.

Divisibility by 10

Rule: A number is divisible by 10 if its last digit is 0.

For example, 450 ends in 0, so it is divisible by 10. However, 47 ends in 7, so it is not divisible by 10.

Any number that ends in 0 is a multiple of 10 because it can be expressed as  N=10⋅(some integer).

Divisibility by 11

Rule: Alternate adding and subtracting the digits from left to right. If the result is divisible by 11 (or is 0), then the original number is divisible by 11.

For example, for 2728, alternate the sum: 2 - 7 + 2 - 8 = -11. 

Since -11 is divisible by 11, 2728 is divisible by 11.

Let's understand the reasoning behind this trick.

Any decimal number N can be represented as : 

N = a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d​, where a, b, c, d are the digits of the numbers.

So if we observe a property, i.e. when you divide the power of 10 by 11 

i.e 10%3 = -1, 

But when 100%11 = 1, 

and then when 1000%11 = -1 and so on.

So for any integer n, 10^n % 11 = -1 if n is odd, and - if n is even

Now so, N = a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d, if we take the modulo of N with 11 we get,

N%11 = (a​*10^k + b*10^(k−1) + ⋯ + c​*10^1 + d)%11.

= ((a​*10^k)%11 + (b*10^(k−1))%11 + ⋯ + (c​*10^1)%11 + d%11)

N%11 = (a-b+c-d​) % 11.

The alternating sum a-b+c-d+e….. and so on effectively gives the remainder when N is divided by 11. 

Therefore, if this alternating sum is divisible by 11 (or is zero), then the original number N is divisible by 11.

Divisibility by 12

Rule: A number is divisible by 12 if it is divisible by both 3 and 4.
So we need to check:

  • If the sum of digits divisible by 3
  • If the last two digits divisible by 4

Let’s Understand the Reason Behind this Trick

The number 12 is the product of 3 × 4, and these two numbers are co-prime (they don’t share any common factor except 1).

So, for a number to be divisible by 12, it must satisfy both of the following:

  • Be divisible by 3 → Check if the sum of its digits is divisible by 3
  • Be divisible by 4 → Check if the last two digits form a number divisible by 4

We don’t need to invent a new rule for 12 because the rules for 3 and 4 are simple, independent, and efficient.
Hence, combining both checks gives us a reliable way to test divisibility by 12.

Example:
Let's take 624.

  • Sum of digits = 6 + 2 + 4 = 12, divisible by 3
  • Last two digits = 24, divisible by 4

So, 624 is divisible by 12.

Divisibility by 13

We don’t have a single straightforward rule for 13. Instead, there are multiple rules or tricks that can help us test divisibility by 13.

Rule 1: Multiply Last Digit by 9 and Subtract
A number is divisible by 13 if you multiply the last digit by 9 and subtract it from the rest of the number, and the result is divisible by 13.

Explanation

This method is based on modular arithmetic, which is a system of arithmetic for integers, where numbers "wrap around" after reaching a certain value, in this case, 13.

Now let’s break this idea into simpler parts:

  • Every number in our decimal system can be broken into two parts:
    • The last digit, say d
    • The rest of the number, say R
      So a number like 78 can be thought of as R = 7, d = 8
  • Mathematically, we write any number N as:
    N = 10 × R + d
    This just means: 10 times the part before the last digit + the last digit
  • But since we’re checking for divisibility by 13, we care about how this expression behaves modulo 13.

Now, here’s the key trick:
We find that 10 ≡ -3 mod 13, which means multiplying by 10 is the same as multiplying by -3 (when we care only about divisibility by 13).

So:

N = 10 × R + d ≡ -3 × R + d (mod 13)

But we want to work with R (the rest of the number), so we rearrange:

N ≡ R - 9 × d (mod 13)

That means: Instead of checking the full number, we can just check if (R - 9 × d) is divisible by 13.

Example:
Take 1,144

  • Last digit = 4
  • Multiply by 9 → 4 × 9 = 36
  • Subtract from rest → 114 - 36 = 78
  • Is 78 divisible by 13? Yes (13 × 6 = 78)

So, 1,144 is divisible by 13

Rule 2: Multiply Last Digit by 4 and Add
A number is divisible by 13 if you multiply the last digit by 4 and add it to the rest of the number, and the result is divisible by 13.

Explanation

This method is again based on modular arithmetic, a way of handling numbers where we’re only interested in remainders after division — in this case, division by 13.

Let’s break this idea into smaller parts:

  1. Every number in base-10 can be broken into:
  • d = the last digit
  • R = the rest of the number (the number without its last digit)

For example, for the number 78:

  • Last digit d = 8
  • Remaining part R = 7

So mathematically, we express this number as:
N = 10 × R + d

This is just how the decimal system works — 78 = (10 × 7) + 8

  1. Now let’s consider this modulo 13:

We’re interested in whether N is divisible by 13, i.e., whether:

N ≡ 0 (mod 13)

So we substitute in our expression:
N = 10 × R + d ≡ ? (mod 13)

  1. Now the trick:

We know from modular arithmetic that:
10 ≡ -9 (mod 13)
(Because 10 and -9 leave the same remainder when divided by 13)

So we can replace 10 × R with (-9 × R):

N = 10 × R + d ≡ -9 × R + d (mod 13)

But this is hard to work with since the variable R is multiplied.
So instead, let’s take another route — we shift the transformation onto the digit instead.

Let’s isolate R:

We want a form like:
N ≡ R + something_with_d (mod 13)

To do that, we observe another equivalent identity:

We find through derivation or pattern experimentation that:
N ≡ R + 4 × d (mod 13)

So this gives us a clean and usable rule:

Example:
Take 351

  • Last digit = 1
  • Multiply by 4 → 1 × 4 = 4
  • Add to rest → 35 + 4 = 39
  • Is 39 divisible by 13? Yes (13 × 3 = 39)

So, 351 is divisible by 13

Rule 3: Multiply Last Digit by 16 and Subtract
A number is divisible by 13 if you multiply the last digit by 16 and subtract it from the rest of the number, and repeat the process until it’s small enough to check.

Explanation

We're interested in checking if the number is divisible by 13, so we want to evaluate:

N = 10 × R + d mod 13

We know from number theory:

10 ≡ -3 (mod 13)
So:
N ≡ 10 × R + d
≡ (-3 × R) + d
≡ d - 3R (mod 13)

This is a valid form, but we want a simpler trick involving subtraction. Let’s go further.

Now instead of stopping at -3, we can try to adjust the values so that the subtraction step becomes easier.

So we try to rearrange the congruence into a subtraction form where:

N ≡ R - 16 × d (mod 13)

How does this work?
Well, it’s based on the fact that:

10 × R + d ≡ R - 16 × d (mod 13)
This is derived by tweaking the multiplication weights until we land on an equivalent form that makes subtraction possible.

So, instead of checking the full number, we can just check if:
R - 16 × d is divisible by 13

Example:
Take 845

  • Last digit = 5
  • Multiply by 16 → 5 × 16 = 80
  • Subtract → 84 - 80 = 4
  • 4 is not divisible by 13 → So, 845 is not divisible by 13

Divisibility by 14

Rule: A number is divisible by 14 if it is divisible by both 2 and 7.
So we need to check:

  • If the last digit is even (i.e., divisible by 2)
  • If the number satisfies the rule for 7. Subtract twice the last digit from the rest of the number. If the result is divisible by 7, so is the original number.

Detailed Explanation

The number 14 is the product of 2 × 7, and both of these numbers are co-prime (they share no common factors other than 1).

So, for a number to be divisible by 14, it must meet both conditions:

  • It should be divisible by 2 → which we check by looking at the last digit (it must be even: 0, 2, 4, 6, or 8)
  • It should be divisible by 7 → which we can check using a known rule:
    Subtract 2 × (last digit) from the rest of the number. If the result is divisible by 7, then so is the original number.

Since both checks are simple and independent, we don’t need a separate rule for 14.

Example: Take 672

  • Last digit = 2 → even
  • Apply rule for 7: 67 - (2 × 2) = 67 - 4 = 63
  • 63 ÷ 7 = 9 → divisible by 7
    So, 672 is divisible by 14

Divisibility by 15

Rule: A number is divisible by 15 if it is divisible by 3 and 5.

  • Check if sum of digits is divisible by 3
  • Check if last digit is 0 or 5

Detailed Explanation

The number 15 is the product of 3 × 5, both of which are co-prime (they have no common factors other than 1).
So for a number to be divisible by 15, it must satisfy both rules:

  • Be divisible by 3 → Check using sum of digits
  • Be divisible by 5 → Check using last digit (0 or 5)

We don’t need to come up with a new rule from scratch because the individual rules for 3 and 5 are simple, independent, and efficient.
Hence, combining both checks gives us a reliable way to test for divisibility by 15.

Example:
Take 345:

  • 3 + 4 + 5 = 12 → divisible by 3
  • Last digit is 5
    So, 345 is divisible by 15.

Divisibility by 16

Rule: A number is divisible by 16 if the last four digits (or the entire number if it has four digits or fewer) form a number that is divisible by 16.

Detailed Explanation

The number 16 is a power of 2 (specifically, 2^4), which means it deals with binary divisibility.
Because 10000 (which is 10^4) is divisible by 16, the last four digits of any number determine its divisibility by 16 — just like how the last two digits matter for divisibility by 4.
So, we only need to check the last 4 digits of the number. If that 4-digit number is divisible by 16, then the entire number is divisible by 16.

Example:
Take 7,456

  • Last 4 digits = 7456
  • Now check: Is 7456 ÷ 16 an integer?

Let's divide:
7456 ÷ 16 = 466, which is a whole number

So, 7456 is divisible by 16

Divisibility by 17

Rule: A number is divisible by 17 if you multiply the last digit by 5, subtract it from the rest of the number, and the result is divisible by 17.
This rule can be applied repeatedly until the number is small enough to check directly.

Detailed Explanation

This trick is based on modular arithmetic, which focuses on how numbers behave when divided by a fixed number — in this case, 17.

Let’s say we have a number N, like 867. We can split it into two parts:

  • The last digit d = 7
  • The rest of the number R = 86

We can always express any number as:
N = 10 × R + d

But since we’re testing divisibility by 17, we look at this expression modulo 17.

To simplify it, we figure out how 10 behaves mod 17:
10 ≡ -7 mod 17

So we substitute back:
N = 10 × R + d ≡ -7 × R + d (mod 17)

Now we rearrange this to work with R:
N ≡ R - 5 × d (mod 17)

This means: Instead of checking the full number N, we can just check if R - 5 × d is divisible by 17.

Example
Take 867.

  • Last digit d = 7
  • Rest of the number R = 86
  • Compute: 86 - 5 × 7 = 86 - 35 = 51

Now check if 51 is divisible by 17:

  • 17 × 3 = 51

So, 867 is divisible by 17.

Divisibility by 18

Rule: A number is divisible by 18 if it is divisible by 2 and 9.
So we need to check:

  • If last digit is even, and
  • If sum of digits is divisible by 9

Detailed Explanation

The number 18 is the product of 2 × 9, and these two numbers are co-prime (they don’t share any common factor except 1).

So, if a number is divisible by both 2 and 9, then it's automatically divisible by 18.

  • The rule for 2 is simple: just check if the last digit is even (i.e., ends in 0, 2, 4, 6, or 8).
  • The rule for 9: check if the sum of all digits is divisible by 9.

We don’t need to come up with a new rule from scratch because the individual rules for 2 and 9 are simple, independent, and efficient.
Hence, combining both checks gives us a reliable way to test for divisibility by 15.

Example:
Take 1,458

  • Last digit is 8 → even
  • Sum of digits: 1 + 4 + 5 + 8 = 18 → divisible by 9
    So, 1,458 is divisible by 18.

Divisibility by 19

Rule: A number is divisible by 19 if you multiply the last digit by 2, subtract it from the rest of the number, and the result is divisible by 19.
This rule can be applied repeatedly until the number becomes small enough to check directly.

Detailed Explanation

This method is based on modular arithmetic, which helps us analyze how numbers behave when divided by a fixed number — in this case, 19.

Suppose we have a number N, like 133. We break it into two parts:

  • The last digit d = 3
  • The rest of the number R = 13

We can always express any number in this form:
N = 10 × R + d

Now, since we're checking divisibility by 19, we'll look at how this behaves modulo 19.

We ask: what is 10 mod 19?
We find that:
10 ≡ -9 mod 19

Now plug this back into the expression:
N = 10 × R + d ≡ -9 × R + d (mod 19)

We rearrange the expression to isolate R:
N ≡ R - 2 × d (mod 19)

So, instead of checking the full number N, we can simply check if R - 2 × d is divisible by 19.

Example:
Take 133

  • Last digit d = 3
  • Rest of the number R = 13
  • Compute: 13 - 2 × 3 = 13 - 6 = 7

Now check if 7 is divisible by 19 → It’s not, so 133 is not divisible by 19.

Let’s try another one:

Example 2:
Take 285

  • Last digit d = 5
  • Rest of the number R = 28
  • Compute: 28 - 2 × 5 = 28 - 10 = 18

Now check if 18 is divisible by 19 → Still not. But we repeat the process.

Now apply the rule again on 18:

  • Last digit d = 8
  • Rest R = 1
  • Compute: 1 - 2 × 8 = 1 - 16 = -15

Since -15 is not divisible by 19, 285 is not divisible by 19.

But if we get 0 or any multiple of 19, that confirms divisibility.

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Divisibility by 20

Rule: A number is divisible by 20 if it ends in 00, 20, 40, 60, or 80.

Detailed Explanation

The number 20 is the product of 4 × 5, and these two numbers are co-prime.

So, for a number to be divisible by 20, it must be divisible by both:

  • 4 → A number is divisible by 4 if its last two digits form a number divisible by 4
  • 5 → A number is divisible by 5 if its last digit is 0 or 5

Now, if a number is divisible by both 4 and 5, it must end in 0 (since divisible by 5) and the last two digits must be divisible by 4.

So effectively, the number must end in 00, 20, 40, 60, or 80, which are all multiples of 20.

Example:
Take 2,140 → ends in 40, and 40 is divisible by 20
So, 2,140 is divisible by 20.

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