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

Playing with Numbers

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) :

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 3

Solution:

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers 4

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

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