## MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.3

**MP Board Class 6 Maths Solution Chapter 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 11 (say, yes or no):

Solution:

**Class 6 Maths Chapter 3 MP Board 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:

(a) 572 is divisible by 4 as its last two digits are divisible by 4, but it is not divisible

by 8 as its last three digits are not divisible by 8.

(b) 726352 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(c) 5500 is divisible by 4 as its last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(d) 6000 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(e) 12159 is not divisible by 4 and 8 as it is an odd number.

(f) 14560 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(g) 21084 is divisible by 4 as its last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(h) 31795072 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(i) 1700 is divisible by 4 as its last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(j) 2150 is not divisible by 4 as its last two digits are not divisible by 4 and it is not divisible by 8 as its last three digits are not divisible by 8.

**MP Board Class 6th Maths Chapter 3 Question 3.**

Using divisibility tests, determine which of following numbers are divisible by 6:

(a) 297144

(c) 4335

(e) 901352

(f) 438750

(g) 1790184

(h) 12583

(i) 639210

(j) 17852

Solution:

(a) 297144 is divisible by 2 as its ones place is an even number place is an even number and it is also divisible by 3 as sum of its digits (= 27) is divisible by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(b) 1258 is divisible by 2 as its ones place is an even number, but it is not divisible by 3 as sum of its digits (= 16) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore, it is not divisible by 6.

(c) 4335 is not divisible by 2 as its ones place is not an even number, but it is divisible by 3 as sum of its digits (= 15) is divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore, it is not divisible by 6.

(d) 61233 is not divisible by 2 as its ones place is not an even number, but it is divisible by 3 as sum of its digits (= 15) is divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore, it is not divisible by 6.

(e) 901352 is divisible by 2 as its ones place is an even number, but it is not divisible by 3 as sum of its digits (= 20) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore, it is not divisible by 6.

(f) 438750 is divisible by 2 as its ones place

is an even number and it is also divisible by 3 as sum of its digits (= 27) is divisible by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(g) 1790184 is divisible by 2 as its ones place is an even number and it is also divisible by 3 as sum of its digits (= 30) is divisible by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(h) 12583 is not divisible by 2 as its ones place is not an even number and it is also not divisible by 3 as sum of its digits (= 19) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore, it is not divisible by 6.

(i) 639210 is divisible by 2 as its ones place is an even number and it is also divisible by 3 as sum of its digits (= 21) is divisible by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is divisible by 6.

(j) 17852 is divisible by 2 as its ones place is an even number, but it is not divisible by 3 as sum of its digits (= 23) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore, it is not divisible by 6.

**Class 6th Maths Chapter 3 Exercise 3.3 : Question 3 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:

(a) In 5445, sum of the digits at odd places = 5 + 4 = 9

Sum of the digits at even places = 4 + 5 = 9

Difference of both sums = 9 – 9 = 0

Since the difference is 0. Therefore, the number is divisible by 11.

(b) In 10824, sum of the digits at odd places = 4 + 8 + 1 = 13

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

Difference of both sums = 13 – 2 = 11

Since the difference is divisible by 11.

Therefore, the number is divisible by 11.

(c) In 7138965, sum of the digits at odd places = 5 + 9 + 3 + 7 = 24

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

Difference of both sums = 24 – 15 = 9

Since the difference is neither 0 nor divisible by 11. Therefore, the number is not divisible by 11.

(d) In 70169308, sum of the digits at odd places = 8 + 3 + 6 + 0 = 17

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

Difference of both sums = 17 – 17 = 0

Since the difference is 0. Therefore, the number is divisible by 11.

(e) In 10000001, sum of the digits at odd places = 1 + 0 + 0 + 0 = 1

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

Difference of both sums = 1 – 1 = 0

Since the difference is 0. Therefore, the number is divisible by 11.

(f) In 901153, sum of the digits at odd places =3 + 1 + 0 = 4

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

Difference of both sums = 15 – 4 = 11

Since the difference is 11. Therefore, the number is divisible by 11.

**6th Class Maths Chapter 3 Exercise 3.3 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 is divisible by 3 if the sum of all digits is divisible by 3.

(a) The smallest digit will be 2.

∴ The number formed is 26724 and

2 + 6 + 7 + 2 + 4 = 21, which is divisible by 3.

And the greatest digit will be 8.

∴ The number formed is 86724 and

8 + 6 + 7 + 2 + 4 = 27, which is divisible by 3.

(b) The smallest digit will be 0.

∴ The number formed is 476502 and

4 + 7 + 6 + 5 + 0 + 2 = 24, which is divisible by 3.

And the greatest digit will be 9.

∴ The number formed is 476592 and

4 + 7 + 6 + 5 + 9 + 2 = 33, which is divisible by 3.

**Exercise 3.3 Class 6 Maths 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:

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

The number formed is 928389

Sum of digits at odd places = 9 + 3 + 2 = 14

Sum of digits at even places = 8 + 8 + 9 = 25

Their difference = 25 – 14 = 11, which is divisible by 11.

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

The number formed is 869484 Sum of digits at odd places = 4 + 4 + 6 = 14

Sum of digits at even places = 8 + 9 + 8 = 25

Their difference = 25 – 14 = 11, which is divisible by 11.