NCERT Solutions for Class 6 Maths Exercise 3.3

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 NCERT Solutions for Class 6 Maths Playing with numbers

Class –VI

Mathematics

(Ex. 3.3)

Question 1.Using divisibility test, 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)

Number	Divisible by
128 990 1586 275 6686 639210 429714 2856 3060 406839	Yes	No	Yes	No	No	Yes	No	No	No

Answer: Sol.

Number	Divisible by
	2	3	4	5	6	8	9	10	11
128 990 1586 275 6686 639210 429714 2856 3060 406839	Yes Yes Yes No Yes Yes Yes Yes Yes No	No Yes No No No Yes Yes Yes Yes Yes	Yes No No No No No No Yes Yes No	No Yes No Yes No Yes No No Yes No	No Yes No No No Yes Yes Yes Yes No	Yes No No No No No No Yes No No	No Yes No No No No Yes No Yes No	No Yes No No No Yes No No Yes No	No Yes No Yes No Yes No No No No

NCERT Solutions for Class 6 Maths Exercise 3.3

Question 2.Using divisibility test, determine which of the following numbers are divisibly by 4; by 8:

(a) 572, (b) 726352, (c) 5500, (d) 6000, (e) 12159, (f) 14560, (g) 21084, (h) 31795072, (i) 1700, (j) 2150

Answer:

(a) 572 {tex} \to {/tex} Divisible by 4 as its last two digits are divisible by 4.

{tex} \to {/tex} Not divisible by 8 as its last three digits are not divisible by 8.

(b) 726352 {tex} \to {/tex} Divisible by 4 as its last two digits are divisible by 4.

{tex} \to {/tex} Divisible by 8 as its last three digits are divisible by 8.

(c) 5500 {tex} \to {/tex} Divisible by 4 as its last two digits are divisible by 4.

{tex} \to {/tex} Not divisible by 8 as its last three digits are not divisible by 8.

(d) 6000 {tex} \to {/tex} Divisible by 4 as its last two digits are 0.

{tex} \to {/tex} Divisible by 8 as its last three digits are 0.

(e) 12159 {tex} \to {/tex} Not divisible by 4 and 8 as it is an odd number.

(f) 14560 {tex} \to {/tex} Divisible by 4 as its last two digits are divisible by 4.

{tex} \to {/tex} Divisible by 8 as its last three digits are divisible by 8.

(g) 21084 {tex} \to {/tex} Divisible by 4 as its last two digits are divisible by 4.

{tex} \to {/tex} Not divisible by 8 as its last three digits are not divisible by 8.

(h) 31795072 {tex} \to {/tex} Divisible by 4 as its last two digits are divisible by 4.

{tex} \to {/tex} Divisible by 8 as its last three digits are divisible by 8.

(i) 1700 {tex} \to {/tex} Divisible by 4 as its last two digits are 0.

{tex} \to {/tex} Not divisible by 8 as its last three digits are not divisible by 8.

(j) 5500 {tex} \to {/tex} Not divisible by 4 as its last two digits are not divisible by 4.

{tex} \to {/tex} Not divisible by 8 as its last three digits are not divisible by 8.

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Question 3.Using divisibility test, 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

Answer:

(a) 297144 {tex} \to {/tex} Divisible by 2 as its units place is an even number.

{tex} \to {/tex} 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 {tex} \to {/tex} Divisible by 2 as its units place is an even number.

{tex} \to {/tex} 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 {tex} \to {/tex} Not divisible by 2 as its units place is not an even number.

{tex} \to {/tex} 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 {tex} \to {/tex} Not divisible by 2 as its units place is not an even number.

{tex} \to {/tex} 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 {tex} \to {/tex} Divisible by 2 as its units place is an even number.

{tex} \to {/tex} 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 {tex} \to {/tex} Divisible by 2 as its units place is an even number.

{tex} \to {/tex} Divisible by 3 as sum of its digits (= 27) is not divisible by 3.

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

(g) 1790184 {tex} \to {/tex} Divisible by 2 as its units place is an even number.

{tex} \to {/tex} Divisible by 3 as sum of its digits (= 30) is not divisible by 3.

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

(h) 12583 {tex} \to {/tex} Not divisible by 2 as its units place is not an even number.

{tex} \to {/tex} 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 {tex} \to {/tex} Divisible by 2 as its units place is an even number.

{tex} \to {/tex} Divisible by 3 as sum of its digits (= 21) is not divisible by 3.

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

(j) 17852 {tex} \to {/tex} Divisible by 2 as its units place is an even number.

{tex} \to {/tex} 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.

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NCERT Solutions for Class 6 Maths Exercise 3.3

Question 4.Using divisibility test, determine which of the following numbers are divisible by 11:

(a) 5445, (b) 10824, (c) 7138965, (d) 70169308, (e) 10000001 , (f) 901153

Answer: (a) 5445 {tex} \to {/tex} Sum of the digits at odd places = 4 + 5 = 9

{tex} \to {/tex} Sum of the digits at even places = 4 + 5 = 9

{tex} \to {/tex} Difference of both sums = 9 – 9 = 0

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

(b) 10824 {tex} \to {/tex} Sum of the digits at odd places = 4 + 8 +1 = 13

{tex} \to {/tex} Sum of the digits at even places = 2 + 0 = 2

{tex} \to {/tex} Difference of both sums = 13 – 2 = 11

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

(c) 7138965 {tex} \to {/tex} Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24

{tex} \to {/tex} Sum of the digits at even places = 6 + 8 + 1 = 15

{tex} \to {/tex} Difference of both sums = 24 – 15 = 9

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

(d) 70169308 {tex} \to {/tex} Sum of the digits at odd places = 8 + 3 + 6 + 0 = 17

{tex} \to {/tex} Sum of the digits at even places = 0 + 9 + 1 + 7 = 17

{tex} \to {/tex} Difference of both sums = 17 – 17 = 0

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

(e) 10000001 {tex} \to {/tex} Sum of the digits at odd places = 1 + 0 + 0 + 0 = 1

{tex} \to {/tex} Sum of the digits at even places = 0 + 0 + 0 + 1 = 1

{tex} \to {/tex} Difference of both sums = 1 – 1 = 0

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

(f) 901153 {tex} \to {/tex} Sum of the digits at odd places = 3 + 1 + 0 = 4

{tex} \to {/tex} Sum of the digits at even places = 5 + 1 + 9 = 15

{tex} \to {/tex} Difference of both sums = 15 – 4 = 11

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

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Question 5.Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisibly by 3:

(a) __________ 6724

(b) 4765 __________ 2

Answer:

(a) We know that a number is divisible by 3 if the sum of all digits is divisible by 3.

Therefore, Smallest digit : 2 {tex} \to {/tex} 26724 = 2 + 6 + 7 + 2 + 4 = 21

Largest digit : 8 {tex} \to {/tex} 86724 = 8 + 6 + 7 + 2 + 4 = 27

(b) We know that a number is divisible by 3 if the sum of all digits is divisible by 3.

Therefore, Smallest digit : 0 {tex} \to {/tex} 476502 = 4 + 7 + 6 + 5 + 0 + 2 = 24

Largest digit : 9 {tex} \to {/tex} 476592 = 4 + 7 + 6 + 5 + 0 + 2 = 33

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NCERT Solutions for Class 6 Maths Exercise 3.3

Question 6.Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisibly by 11:

(a) 92 __________ 389

(b) 8 __________ 9484

Answer: (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 should be either 0 or 11.

Therefore, 928389 {tex} \to {/tex} Odd places = 9 + 8 + 8 = 25

Even places = 2 + 3 + 9 = 14

Difference = 25 – 14 = 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 should be either 0 or 11.

Therefore, 869484 {tex} \to {/tex} Odd places = 8 + 9 + 8 = 25

Even places = 6 + 4 + 4 = 14

Difference = 25 – 14 = 11

NCERT Solutions for Class 6 Maths Exercise 3.3

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