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# Divisibility Rules

• Divisibility rules are essential to know that a number is divisible by another number or not.
• With the help of these rules, we can identify the factors of a number.
• Every multiple of a number is divisible by that number.

For example:

4, 6, 8,.... are multiples of 2, so, they are all divisible by 2.

• For one digit numbers and small numbers of two digits, we can easily figure out by which number it is divisible.

For example:

For numbers like, 2, 4, 9, 12 etc, it is easy to know that by which numbers they are divisible.

• But for large numbers, it is not so easy.
• To solve these types of problems, we need divisibility rules.

### Divisibility by 2

• A number is divisible by 2 if the last digit is zero or an even number like 2, 4, 6 or 8.
• All even numbers are divisible by 2 because they are all multiples of 2 and zero is also divisible by 2.

Examples:

• Consider the number 246 and notice its last digit which is 6. Here 6 is an even number. So, it is divisible by 2.
• Number 250 is divisible by 2 as its last digit is zero which is divisible by 2.
• Number 145 is not divisible by 2 as its last digit is 5 which is not an even number.

#### Which one of the following is NOT divisible by 2?

A 240

B 125

C 122

D 236

×

In the number 240, the last digit is zero, which is divisible by 2. So, the whole number 240 is divisible by 2.

In the number 125, the last digit is 5, which is an odd number. So, the whole number 125 is not divisible by 2.

In the number 122, the last digit is 2, which is an even number. So, the whole number 122 is divisible by 2.

In the number 236, the last digit is 6, which is an even number. So, the whole number 236 is divisible by 2.

Hence, option (B) is correct.

### Which one of the following is NOT divisible by 2?

A

240

.

B

125

C

122

D

236

Option B is Correct

# Common Multiple

A multiple that is common in two or more numbers is called common multiple.

For Example:  Here, $$6,\, 12,\, 18\,...$$ are common among the multiples of both 2 and 3.

Thus, the common multiples of 2 and 3 $$=6,\, 12,\, 18\,...$$

#### Which one of the following is a common multiple of 5 and 6?

A 25

B 36

C 30

D 12

× Here, $$30$$ is common among the multiples of both 5 and 6.

Thus, the common multiple of 5 and 6 is $$30$$.

Hence, option (C) is correct.

### Which one of the following is a common multiple of 5 and 6?

A

25

.

B

36

C

30

D

12

Option C is Correct

# Divisibility Rule of 3

A number is divisible by 3 if the sum of all the digits is a multiple of 3.

For example:

1. Consider the number: 2451

The sum of all the digits of the number = 2 + 4 + 5 + 1 = 12

• The sum is 12, which is a multiple of 3. So, the number 2451 is divisible by 3.

2. Consider another number: 5324

The sum of all the digits of the number = 5 + 3 + 2 + 4 = 14

• The sum is 14, which is not a multiple of 3. So, the number 5324 is not divisible by 3.

Case:

• If the number has too many digits then there is quite a possibility that the sum of all the digits is large.
• In that case, take that sum as a new number and repeat the process.

For example:

1. Consider the number: 996897

The sum of all the digits = 9 + 9 + 6 + 8 + 9 + 7  = 48

48 is also a big number, so we can take it as a new number and then repeat the process.

48 = New number

The sum of all the digits = 4 + 8 = 12

12 is a multiple of 3, so the original number is also divisible by 3.

2. Consider the number: 6896756

The sum of all the digits = 6 + 8 + 9 + 6 + 7 + 5 + 6 = 47

Take 47 as a new number and repeat the process.

47 = New number

The sum of all the digits = 4 + 7 = 11

11 is not a multiple of 3, so the original number is not divisible by 3.

#### Which one of the following numbers is divisible by 3?

A 23

B 3513

C 5921

D 4805

×

For the number 23:

The sum of all digits = 2 + 3  = 5

5 is not a multiple of 3, so the number 23 is not divisible by 3.

For the number 3513:

The sum of all digits = 3 + 5 + 1 + 3 = 12

12 is a multiple of 3, so the number 3513 is divisible by 3.

For the number 5921:

The sum of all digits = 5 + 9 + 2 + 1 = 17

17 is not a multiple of 3, so the number 5921 is not divisible by 3.

For the number 4805:

The sum of all digits = 4 + 8 + 0 + 5 = 17

17 is not a multiple of 3, so the number 4805 is not divisible by 3.

Hence, option (B) is correct.

### Which one of the following numbers is divisible by 3?

A

23

.

B

3513

C

5921

D

4805

Option B is Correct

# Multiple

• A multiple is a product of a quantity and a whole number.

For example:

Multiples of $$2$$ can be represented by a product of $$2$$ and the whole numbers.

Multiples of $$2=2×\text{whole numbers}$$

$$1^{st}$$ Multiple of $$2=2×1=2$$

$$2^{nd}$$ Multiple of $$2=2×2=4$$

$$3^{rd}$$ Multiple of $$2=2×3=6$$

$$4^{th}$$ Multiple of $$2=2×4=8$$

$$5^{th}$$ Multiple of $$2=2×5=10$$

and so on.

• Thus, the multiples of $$2$$ are $$2,\;4,\;6,\;8,\;10\,...$$

#### Find the first five multiples of $$5.$$

A $$5,\;10,\;15,\;20,\;25$$

B $$5,\;10,\;15,\;20,\;30$$

C $$10,\;15,\;20,\;25,\;30$$

D $$10,\;20,\;30,\;40,\;50$$

×

The multiples of $$5$$ can be represented by

Multiples of $$5=5×\text{whole numbers}$$

$$1^{st}$$ Multiple of $$5=5×1=5$$

$$2^{nd}$$ Multiple of $$5=5×2=10$$

$$3^{rd}$$ Multiple of $$5=5×3=15$$

$$4^{th}$$ Multiple of $$5=5×4=20$$

$$5^{th}$$ Multiple of $$5=5×5=25$$

Thus, the first five multiples of $$5$$ are $$5,\;10,\;15,\;20$$ and $$25$$.

Hence, option (A) is correct.

### Find the first five multiples of $$5.$$

A

$$5,\;10,\;15,\;20,\;25$$

.

B

$$5,\;10,\;15,\;20,\;30$$

C

$$10,\;15,\;20,\;25,\;30$$

D

$$10,\;20,\;30,\;40,\;50$$

Option A is Correct

# Divisibility Rule of 4

A number is divisible by 4 if

• Case I:
• The last two digits of the number are a multiple of 4.
• For example:
• Consider the number 5824, the last two digits are 24 which is a multiple of 4, so the number 5824 is divisible by 4.
• Consider another number 5825, the last two digits of this number are 25 which is not a multiple of 4, so 5825 is not divisible by 4.
• Case II:
• A number is divisible by 4 if the last two digits of that number are 00.
• For example:
• Consider the number 3200, the last two digits are 00. So, the number is divisible by 4.

#### Which one of the following numbers is divisible by 4?

A 402

B 232

C 242

D 223

×

In the number 402, the last two digits are 02, and 02 is not a multiple of 4, so the number 402 is not divisible by 4.

In the number 232, the last two digits are 32, and 32 is a multiple of 4, so the number 232 is divisible by 4.

In the number 242, the last two digits are 42, and 42 is not a multiple of 4, so the number 242 is not divisible by 4.

In the number 223, the last two digits are 23, and 23 is not a multiple of 4, so the number 223 is not divisible by 4.

Hence, option (B) is correct.

### Which one of the following numbers is divisible by 4?

A

402

.

B

232

C

242

D

223

Option B is Correct

# Divisibility Rule of 5

• A number is divisible by 5 if the last digit of the number is either zero or 5.

For example:

• Consider the three numbers, 515, 420 and 226.
• In 515, the last digit is 5, so 515 is divisible by 5.
• In 420, the last digit is zero, so 420 is divisible by 5.
• In 226, the last digit is neither 5 nor zero, so 226 is not divisible by 5.

#### Which one of the following numbers is divisible by 5?

A 415

B 4038

C 328

D 312

×

In the number 415, the last digit is 5, so 415 is divisible by 5.

In the number 4038, the last digit is neither 5 nor zero, so 4038 is not divisible by 5.

In the number 328, the last digit is neither 5 nor zero, so 328 is not divisible by 5.

In the number 312, the last digit is neither 5 nor zero, so 312 is not divisible by 5.

Hence, option (A) is correct.

### Which one of the following numbers is divisible by 5?

A

415

.

B

4038

C

328

D

312

Option A is Correct