# Concept: Divisibility Rules

CONTENTS

**Divisibility Rule for 2**

^{n}For N to be completely divisible by a, remainder when N is divided by a should be 0.

**Divisibility Rule for 2**

^{n}For a number to be divisible by 2^{n}, the number formed by its last 'n' digits should be divisible by 2^{n}.

**Divisibility Rule for 2**

For a number to be divisible by 2, it's last digit should be divisible by 2.

**Divisibility Rule for 4**

For a number to be divisible by 4, the number formed by its last 2 digits should be divisible by 4.

**Divisibility Rule for 8**

For a number to be divisible by 8, the number formed by its last 3 digits should be divisible by 8.

**Divisibility Rule for 3 or 9**

For a number to be divisible by 3 or 9, sum of it's digits should be divisible by 3 or 9 respectively.

**Divisibility Rule for 5**

^{n}For a number to be divisible by 5^{n}, the number formed by its last 'n' digits should be divisible by 5^{n}.

**Divisibility Rule for 5**

For a number to be divisible by 5, its last digit should be divisible by 5.

**Divisibility Rule for 25**

For a number to be divisible by 25, the number formed by its last 2 digits should be divisible by 25.

**Divisibility Rule for 125**

For a number to be divisible by 125, the number formed by its last 3 digits should be divisible by 125.

**Divisibility Rule for 7**

For a number to be divisible by 7, then the difference between twice the unit's digit of the given number and the remaining part of the given number should be a multiple of 7 or it should be equal to 0.

**Divisibility Rule for 11**

To find if a number is divisible by 11 or not, we need to calculate sum of alternated digits starting with unit;s digit (u) and sum of alternate digits starting with ten's digit (t)

If (u - t) is divisible by 11 (ignoring its sign), then the given number is also divisible by 11.

**Divisibility Rule for any composite number**

To find if a given number 'N' is divisible by 'p' or not, write 'p' as a product of two co-prime numbers, i.e., p = a × b

If 'N' is divisible by both 'a' and 'b', then 'N' is also divisible by 'p'.

- If N is divisible by both 'a' and 'b', then it will also be divisible by LCM(a, b)
- If N is divisible by numbers 'a', 'b', 'c', and so on, then it will also be divisible by LCM(a, b, c, ...)

**Divisibility Rule for 6**

For a number to be divisible by 6 (= 2 × 3), the number should be divisible by both 2 and 3.

**Divisibility Rule for 12**

For a number to be divisible by 12 (= 3 × 4), the number should be divisible by both 3 and 4.

**Divisibility Rule for 36**

For a number to be divisible by 36 (= 4 × 9), the number should be divisible by both 4 and 9.

**Divisibility Rule for 10**

^{n}± 1**Divisibility Rule for 10**

^{n}+ 1This divisiblity Rule is similar to that of 11 (10^{1} + 1). For 11 we took sum of alternate digits taking only one digit at a time since n = 1.

If n = 2, we will take sum of alternate pair digits taking 2 digits at a time.

**Example**: Find if 5741648 is divisible by 101 or not.

**Solution:**

Here we need to find divisiblity with 101 (10^{2} + 1). Hence we make pair of 2-digit starting from right side.

5 74 16 48

Now, we add the alternate pair of digits

x = 48 + 74 = 122

y = 16 + 5 = 21

Now, x - y = 122 - 21 = 101.

Since, x - y is divisible by 101, the original number 5741648 is also divisible by 101.

This concept can be extended to any value of n.

**Divisibility Rule for 10**

^{n}- 1This divisiblity Rule is similar to that of 9 (10^{1} - 1). For 9 we took sum of one digit at a time since n = 1.

If n = 2, we will take sum of digits taking 2 digits at a time.

**Example**: Find if 5741648 is divisible by 99 or not.

**Solution:**

Here we need to find divisiblity with 99 (10^{2} - 1). Hence we make pair of 2-digit starting from right side.

5 74 16 48

Now, we add these pair of digits

5 + 74 + 16 + 48 = 143

Since, 143 is not divisible by 99, the original number 5741648 is also not divisible by 99.

This concept can be extended to any value of n.