NCERT Solutions For Class 6 Maths Chapter 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 11 (say, yes or no):
Answer 1:
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) 317950
(i) 1700 (j) 2150
Answer 2:
If last two digit of number is divisible by 4.then number is divisible by 4.And ,if last three digit of number is divisible by 8.then number is divisible by 8.
(a) 572, It last two digit is 72 which is divisible by 4. So, 572 is divisible by 4.
It last three digit is 572 which is divisible by 8. So, it is divisible by 8.
(b) 726352, It last digit is 52 which is divisible by 4. So, it is divisible by 4.
It last three digit is 572 which is divisible by 8. So, it is divisible by 8.
(c) 5500, It last digit is 00 which is divisible by 4. So, it is divisible by 4.
It last three digit is 572 which is not divisible by 8. So, it is not divisible by 8.
(d) 6000, It last two digit is 00 which is divisible by 4. So, it is divisible by 4.
It last three digit is 000 which is divisible by 8. So, it is divisible by 8.
(e) 12159, It last two digit is 59 which is not divisible by 4. So, it is not divisible by 4.
It last three digit is 159 which is not divisible by 8. So, it is not divisible by 8.
(f) 14560, It last two digit is 60 which is divisible by 4. So, it is divisible by 4.
It last three digit is 560 which is divisible by 8. So, it is divisible by 8.
(g) 21084,It last two digit is 84 which is divisible by 4. So, it is divisible by 4.
It last three digit is 084 which is not divisible by 8. So, it is not divisible by 8.
(h) 31795072, It last two digit is 72 which is divisible by 4. So, it is divisible by 4.
It last three digit is 072 which is divisible by 8. So, it is divisible by 8.
(i) 1700,It last two digit is 00 which is divisible by 4. So, it is divisible by 4.
It last three digit is 700 which is not divisible by 8. So, it is not divisible by 8.
(j) 2150, It last two digit is 50 which is not divisible by 4. So, it is not divisible by 4.
It last three digit is 150 which 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
Answer 3:
We know that if number is divisible by 2 and 3 both then it is divisible by 6 also .
(a) 297144
Its unit’s digit is 4 which is even . So, it is divisible by 2.
Sum of all digits = 2+ 9+ 7 + 1 + 4 + 4 = 27, which is divisible by 3.
So, number is divisible by both 2 and 3 Therefore, 297144 is divisible by 6.
(b) 1258
Its unit’s digit is 8 which is odd. So, it is divisible by 2.
Sum of all digits = 1+ 2 + 5 + 8 = 16, which is not divisible by 3.
Here, number is not divisible by both 2 and 3 Therefore, 1258 is not divisible by 6.
(c) 4335 .
Its unit’s digit is 5 which is odd. So, it is not divisible by 2.
Here, number is not divisible by both 2 and 3 Therefore, 4335 is also not divisible by 6.
(d) 61233
Its unit’s digit is 3 which is odd. So, it is not divisible by 2.
Here, number is not divisible by both 2 and 3 Therefore, 61233 is also not divisible by 6.
(e) 901352
Its unit’s digit is 2 which is even. So, it is divisible by 2.
Sum of its digits = 9+ 0 + 1 + 345 + 2 = 20, which is not divisible by 3.
Here, number is not divisible by both 2 and 3 Therefore 901352 is not divisible by 6.
f) 438750
Its unit’s digit is 0 which is even. So, it is divisible by 2.
Sum of its digits = 4+ 3 + 8 + 7 + 5 + 0=27, which is divisible by 3.
Here, number is divisible by both 2 and 3 Therefore, 438750 is divisible by 6.
(g) 1790184
Its unit’s digit is 4 which is even. So, it is divisible by 2.
Sum of its digits = 1+ 7 + 9+ 0+ 1+ 8 + 4 = 30, which is divisible by 3.
Here, number is divisible by both 2 and 3 Therefore 1790184 is divisible by 6.
(h) Given number = 12583 .
Its unit’s digit is 3 which is odd. So, it is not divisible by 2.
Here, number is not divisible by both 2 and 3 Therefore, 12583 is not divisible by 6.
(i) 639210
Its unit’s digit is 0 which is even. So, it is divisible by 2.
Sum of its digits = 6+ 3+ 9+ 2 + 1 + 0 = 21, which is divisible by 3.
Here, number is divisible by both 2 and 3 Therefore, 639210 is divisible by 6.
(j) Given number = 17852
Its unit’s digit is 2 which is even. So, it is divisible by 2.
Sum of its digits = 1 + 7 + 8 + 5 + 2 = 23, which is not divisible by 3.
Here, number is not divisible by both 2 and 3 Therefore 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
Answer 4:
We know that a number is divisible by 11, if the difference in odd places and the sum of its digits in even places is either 0 or a divisible by 11.
(a) 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 places = 9 – 9 = 0
the difference is 0 5445 is divisible by 11.
(b) 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 places =13 – 2 = 11
the difference is 11 So 10824 is divisible by 11.
(c) 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 places = 24 – 15 = 9
the difference is 9 So 7138965 is not divisible by 11.
(d) 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 places =17 – 17 = 0
the difference is 0 So 70169308 is divisible by 11.
(e) 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
the difference is 0 So 10000001 is divisible by 11.
(f) 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 places =15 – 4 = 11,
the difference is 11 So 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
Answer 5:
We know that if the sum of its digits is divisible by 3 then it divisible by 3,(a) _______6724 Here
6 + 7 + 2 + 4 =19, when we add 2 to 19, the result number 21 will be divisible by 3.
The smallest digit is 2.
Again,
when we add 8 to 19, the result number 27 will be divisible by 3
The largest digit is 8.
(b) 4765 _____ 2 Here
4 + 7 + 6 + 5 + 2 = 24, it is divisible by 3.
the smallest digit is 0.
Again,
when we add 9 to 24, the resulting number 33 will be divisible by 3.
The 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
Answer 6:
We know that a number is divisible by 11, if the difference of the sum of its digits at odd places and even places is either 0 or divisible by 11.(a) 92 ___389
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 complete divisible by 11
=required digit + 3=11
=required digi=11-3=8
So, the required smallest digit = 8 .
(b) 8…9484
sum of digits at even places
= 8 + 9 + 8 = 25
sum of the digits at odd places
= 4 + 4 + required digit
= 8 + required digit
Difference
= 25 – (8 + required digit)
= 17- required digit
or complete divisible by 11
17- required digit=11
=17-11=required digit
So the required smallest digit = 6