NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3.

- Playing with Numbers Class 6 Ex 3.1
- Playing with Numbers Class 6 Ex 3.2
- Playing with Numbers Class 6 Ex 3.4
- Playing with Numbers Class 6 Ex 3.5
- Playing with Numbers Class 6 Ex 3.6
- Playing with Numbers Class 6 Ex 3.7

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Playing with Numbers |

Exercise |
Ex 3.3 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3

**Question 1.**

Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 4; by 11 (say yes or no) :

**Solution:**

**Question 2.**

Using divisibility tests, determine which of the following numbers are divisible by 4; by 8 :

**(a)** 572

**(b)** 726352

**(c)** 5500

**(d)** 6000

**(e)** 12159

**(f)** 14560

**(g)** 21084

**(h)** 31795072

**(i)** 1700

**(j)** 2150

**Solution:**

We know that a number is divisible by 4, if the number formed by its digits in ten’s and unit’s place is divisible by 4.

**(a)** In 572, 72 is divisible by 4. So, 572 is divisible by 4.

**(b)** In 726352, 52 is divisible by 4. So, it is divisible by 4.

**(c)** In 5500, 00 is divisible by 4. So, it is divisible by 4.

**(d)** In 6000, 00 is divisible by 4. So, it is divisible by 4.

**(e)** In 12159, 59 is not divisible by 4. So, it is not divisible by 4.

**(f)** In 14560, 60 is divisible by 4. So, it is divisible by 4.

**(g)** In 21084, 84 is divisible by 4. So, it is divisible by 4.

**(h)** In 31795072, 72 is divisible by 4. So, it is divisible by 4.

**(i)** In 1700,00 is divisible by 4. So, it is divisible by 4.

**(j)** In 2150, 50 is not divisible by 4. So, it is not divisible by 4.

Also, we know that a number is divisible by 8, if the number formed by its hundred’s, ten’s and unit’s places is divisible by 8.

**(a)** 572 is not divisible by 8.

**(b)** In 726352, 352 is divisible by 8. So, it divisible by 8.

**(c)** In 5500, 500 is not divisible by 8. So, it is not divisible by 8.

**(d)** In 6000, 000 is divisible by 8. So, it is divisible by 8.

**(e)** In 12159, 159 is not divisible by 8. So, it is not divisible by 8.

**(f)** In 14560, 560 is divisible by 8. So, it is divisible by 8.

**(g)** In 21084, 084 is not divisible by 8. So, it is not divisible by 8.

**(h)** In 31795072, 072 is divisible by 8. So, it is divisible by 8.

**(i)** In 1700, 700 is not divisible by 8. So, it is not divisible by 8.

**(j)** In 2150, 150 is not divisible by 8. So, it is not divisible by 8.

**Question 3.**

Using divisibility tests, determine which of the following

numbers are divisible by 6 : .

**(a)** 297144

**(b)** 1258

**(c)** 4335

**(d)** 61233

**(e)** 901352

**(f)** 438750

**(g)** 1790184

**(h)** 12583

**(i)** 639210

**(j)** 17852

**Solution:**

We know that a number is divisible by 6, if it is divisible by 2 and 3 both.

**(a)** Given number = 297144

Its unit’s digit is 4. So, it is divisible by 2.

Sum of its digits = 2+ 9+ 7 + 1 + 4 + 4 = 27, which is divisible by 3.

∴ 297144 is divisible by 6.

**(b)** Given number =1258

Its unit’s digit is 8. So, it is divisible by 2.

Sum of its digits = 1+ 2 + 5 + 8 = 16, which is not divisible by 3.

∴ 1258 is not divisible by 6.

**(c)** Given number = 4335 .

Its unit’s digit is 5. So, it is not divisible by 2.

∴ 4335 is also not divisible by 6.

**(d)** Given number = 61233

Its unit’s digit is 3. So, it is not divisible by 2.

∴ 61233 is also not divisible by 6.

**(e)** Given number = 901352

Its unit’s digit is 2. So, it is divisible by 2.

Sum of its digits = 9+ 0 + 1 + 345 + 2 = 20, which is not divisible by 3.

∴ 901352 is not divisible by 6.

**(f)** Given number = 438750

Its unit’s digit is 0. So, it is divisible by 2.

Sum of its digits = 4+ 3 + 8 + 7 + 5 + 0=27, which is divisible by 3.

∴ 438750 is divisible by 6.

**(g)** Given number = 1790184

Its unit’s digit is 4. So, it is divisible by 2.

Sum of its digits = 1+ 7 + 9+ 0+ 1+ 8 + 4 = 30, which is divisible by 3.

∴ 1790184 is divisible by 6.

**(h)** Given number = 12583 .

Its unit’s digit is 3. So, it is not divisible by 2.

∴ 12583 is not divisible by 6.

**(i)** Given number = 639210

Its unit’s digit is 0. So, it is divisible by 2.

Sum of its digits = 6+ 3+ 9+ 2 + 1 + 0 = 21, which is divisible by 3.

∴ 639210 is divisible by 6.

**(j)** Given number = 17852

Its unit’s digit is 2. So, it is divisible by 2.

Sum of its digits = 1 + 7 + 8 + 5 + 2 = 23, which is not divisible by 3.

∴ 17852 is not divisible by 6.

**Question 4.**

Using divisibility tests, determine which of the following numbers are divisible by 11:

**(a)** 5445

**(b)** 10824

**(c)** 7138965

**(d)** 70169308

**(e)** 10000001

**(f)** 901153

**Solution:**

We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.

**(a)** Given number = 5445

Sum of its digits at odd places = 5 + 4 = 9

Sum of its digit at even places = 4 + 5 = 9

Difference of these two sums = 9 – 9 = 0

∴ 5445 is divisible by 11.

**(b)** Given number = 10824

Sum of its digits at odd places = 4+ 8 + 1 = 13

Sum of its digits at even places =2 + 0 =2

Difference of these two sums =13 – 2 = 11, which is a multiple of 11.

∴ 10824 is divisible by 11.

**(c)** Given number = 7138965

Sum of its digits at odd places = 5+ 9+ 3 + 7= 24

Sum of its digits at even places = 6+ 8 + 1 = 15

Difference of these two sums = 24 – 15 = 9, which is not a multiple of 11.

∴ 7138965 is not divisible by 11.

**(d)** Given number = 70169308

Sum of its digits at odd places = 8 + 3 + 6 + 0=17

Sum of its digits at even places = 0 + 9 + 1 + 7 = 17

Difference of these two sums =17 – 17 = 0

∴ 70169308 is divisible by 11.

**(e)** Given number = 10000001

Sum of its digits at odd places = 1 + 0 + 0 + 0 = 1

Sum of its digits at even places = 0 + 0 + 0 + 1 = 1

Difference of these two sums = 1 – 1 = 0

∴ 10000001 is divisible by 11.

**(f)** Given number = 901153

Sum of its digits at odd places = 3 + 1 + 0 = 4

Sum of its digits at even places = 5 + 1 + 9=15

Difference of these two sums =15 – 4 = 11,

which is a multiple of 11.

∴ 901153 is divisible by 11.

**Question 5.**

Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3 :

**(a)** … 6724

**(b)** 4765 … 2

**Solution:**

We know that a number divisible by 3, if the sum of its digits is divisible by 3.

**(a) … 6724**

For … 6724, we have 6 + 7 + 2 + 4 =19, we add 2 to 19, the resulting number 21 will be divisible by 3.

∴ The required smallest digit is 2.

Again, if we add 8 to 19, the resulting number 27 will be divisible by 3. .’. The required largest digit is 8.

**(b) 4765 … 2**

For 4765 … 2, we have 4 + 7 + 6 + 5 + 2 = 24, it is divisible by 3.

Hence the required smallest digit is 0.

Again, if we add 9 to 24, the resulting number 33 will be divisible by 3.

∴ The required largest digit is 9.

**Question 6.**

Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11:

**(a)** 92 … 389

**(b)** 8 … 9484

**Solution:**

We know that a number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or divisible by 11.

**(a) 92 … 389**

For 92 … 389, sum of the digits at odd places and sum of digits at even places

= 9 + 3 + 2 = 14

= 8 + required digit + 9

= required digit+ 17

Difference between these sums

= required digit + 17 – 14

= required digit + 3

For (required digit + 3) to become 11, we must have the required digit as 8 (∵ 3+ 8 gives 11).

Hence, the required smallest digit = 8 .

**(b) 8…9484**

For 8 … 9484, sum of the digits at odd places

= 4 + 4 + required digit

= 8 + required digit

and sum of digits at even places

= 8 + 9 + 8 = 25

Difference between these sums

= 25 – (8 + required digit)

= 17- required digit

For (17 — required digit) to become 11 we must have the required digit as 6 (∵ 17-6 =11).

Hence the required smallest digit = 6

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3, drop a comment below and we will get back to you at the earliest.

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