NCERT Exercise 3.3 Solution Part-4 class six math
In this section question number 4 to 6 of NCERT Exercise 3.3 have been solved. Numbers given in question number 4 are to be tested for divisibility by 11. In question number 5 two numbers have been given in which one digit in each number is missing. The missing number is to be filled in such a way that given numbers will become divisible by 3. In question number 6 two numbers, each with a missing digit, are given. The missing digit is to be identified so that the numbers will become divisible by 11.
Question (4) Using divisibility tests, determine which of the following numbers are divisible by 11:
(a) 5445
Answer
Divisible by 11: Yes
17852 ÷ 11 = 495
Explanation
The difference between the sum of digits at even places and at odd places of 5445 counting from right is equal to zero, thus 5445 is divisible by 11.
= 5 + 4 – (5 + 4)
= 9 – 9 = 0
Thus, 5445 is divisible by 11
17852 ÷ 11 = 495
(b) 10824
Answer
Divisible by 11: Yes
10824 ÷ 11 = 984
Explanation
The difference between the sum of digits at even places and at odd places of 10824 counting from right is equal to eleven (11), thus 10824 is divisible by 11.
= 4 + 8 + 1 – (2 + 0)
= 13 – 2 = 11
Thus, 10824 is divisible by 11
10824 ÷ 11 = 984
(c) 7138965
Answer
Divisible by 11: No
Explanation
The difference between the sum of digits at even places and at odd places of 7138965 counting from right is neither equal to eleven (11) or zero (0), thus 7138965 is not divisible by 11.
= 5 + 9 + 3 + 7 – (6 + 8 + 1)
= 24 – 15 = 9
Thus, 7138965 is not divisible by 11
(d) 70169308
Answer
Divisible by 11: Yes
70169308 ÷ 11 = 6379028
Explanation
The difference between the sum of digits at even places and at odd places of 70169308 counting from right is equal to zero (0), thus 70169308 is divisible by 11.
= 8 + 3 + 6 + 0 – (0 + 9 + 1 + 7)
= 17 – 17 = 0
Thus, 70169308 is divisible by 11
70169308 ÷ 11 = 6379028
(e) 10000001
Answer
Divisible by 11: Yes
10000001 ÷ 11 = 909091
Explanation
The difference between the sum of digits at even places and at odd places of 10000001 counting from right is equal to zero (0), thus 10000001 is divisible by 11.
= 1 + 0 + 0 + 0 – (0 + 0 + 0 + 1)
= 1 – 1 = 0
Thus, 10000001 is divisible by 11
10000001 ÷ 11 = 909091
(f) 901153
Answer
Divisible by 11: Yes
901153 ÷ 11 = 91923
Explanation
The difference between the sum of digits at even places and at odd places of 901153 counting from right is equal to eleven (11), thus 901153 is divisible by 11.
= 5 + 1 + 9 – (3 + 1 + 0)
= 15 – 4 = 11
Thus, 901153 is divisible by 11
901153 ÷ 11 = 81923
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
(a) ___6724
If the sum of the digits of a number is divisible by 3, then the number is divisible by 3 also.
Thus, 2 + 6 + 7 + 2 + 4 = 21
27624 ÷ 3 = 9208
Thus, the smallest number is 2 Answer
(b) 4765__2
4 + 7 + 6 + 5 + 6 + 2 = 30
476562 ÷ 3 = 158854
Thus, the greatest number is 6 Answer
Thus, the smallest number for question (a) = 2 and greatest number for question (b) = 6 Answer
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
If the difference between the sum of digits situated at odd and even places counting from right of a number is either equal to zero (0) or eleven (11), then the number is divisible by 11.
Thus, Answer (a) 92 ____ 389
The sum of digits situated at odd places counting from right
= 9 + 3 + 2 = 14
And, the sum of digits situated at even places counting from right
= 8 + __ + 9
= ___ + 17
Now, 17 – 14 = 3
Here, 11 – 3 = 8
This means if we add 8 to the 3, then it will become 11
Thus, 8 + 8 + 9 = 25
Now, 25 – 14 = 11
This, means if 8 is in blank space then, the given number will be divisible by 11.
Thus, the number is 928389
928389 ÷ 3 = 84399
Thus, the missing number = 8 Answer
Answer (b) 8 __ 9484
The sum of digits situated at odd places counting from right
= 4 + 4 + __ = 8 + __
And, the sum of digits situated at even places counting from right
= 8 + 9 + 8 = 25
Now, 25 – 11 = 14
Thus, missing digit = 14 – 8 = 6
Now, the sum of digits situated at odd places counting from right
= 4 + 4 + 6 = 14
And, the sum of digits situated at even places counting from right
= 8 + 9 + 8 = 25
Now, the difference between digits situated at odd and even places of the given number
= 25 – 14 = 11
Thus, 869484 is divisible by 11
869484 ÷ 3 = 79044
Thus, the missing number = 6 Answer