# DAV Class 6 Maths Chapter 2 Worksheet 3 Solutions

The DAV Class 6 Maths Book Solutions Pdf and DAV Class 6 Maths Chapter 2 Worksheet 3 Solutions of Factors and Multiples offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 2 WS 3 Solutions

Question 1.
Look at the following group of numbers and fill in the blanks:

(a) 389510, 7781450, 4203324, 12342
The numbers divisible by 3 are _______ and _______.
Solution:
4203324, 12342

(b) 3437712, 4222910, 6880172, 9811602
The numbers divisible by 9 are _______ and _______.
Solution:
3437712, 9811602

(c) 362442, 8502153, 774067, 46627207
The numbers divisible by 11 are _______ and _______.
Solution:
8502153, 46627207 Question 2.
Pick out the numbers from the following that are divisible by 3 but not by 9.

(a) 38721
Solution:
38721 is divisible by 3 but not divisible by 9

(b) 422679
Solution:
422679 is divisible by 3 but not divisible by 9

(c) 6110586
Solution:
6110586 is divisible by 3 and 9 both.

(d) 257796
Solution:
257796 is divisible by 3 and 9 both. Question 3.
Test the following for the divisibility by 3 and 9.
(a) 294414
Solution:
294414
Sum of the digits of the number 294414 is
2 + 9 + 4 + 4 + 1 + 4 = 24 + 3 = 8
Hence, 294414 is divisible by 3.

(b) 145404
Solution:
145404
Sum of the digits of the number 145404 is
1+4 + 5 + 4 + 0 + 4=18 + 9 = 2
Hence, 145404 is divisible by 9.

(c) 99999
Solution:
99999
The sum of digits of the number 99999 is
9 + 9 + 9 + 9 + 9 = 45 + 9 = 5
Hence, 99999 is divisible by 9. Question 4.
Test the divisibility of the following numbers by 11:

(a) 86611291
Solution:
86611291
Sum of the digits at odd places is 8 + 6 + 1 + 9 = 24
Sum of the digits at even places is 6 + 1 + 2 + 1 = 10
Difference 24 – 10 = 14 is not divisible by 11
Hence, 86611291 is not divisible by 11.

(b) 100001
Solution:
100001
Sum of the digits at odd places is 1 + 0 + 0 = 1
Sum of the digits at even places is 0 + 0 + 1 = 1
Difference 1 – 1 = 0
Hence, 100001 is divisible by 11.

(c) 9427355
Solution:
9427355
Sum of the digits at odd places is 9 + 2 + 3 + 5 = 19
Sum of the digits at even places is 4 + 7 + 5 = 16
Difference 19 – 16 = 3 is not divisible by 11
Hence, 9427355 is not divisible by 11. (d) 7023643
Solution:
7023643
The sum of the digits at odd places is 9 + 2 + 3 + 5 = 19
Sum of the digits at even places is 4 + 7 + 5 = 16
Difference 19 – 16 = 3 is not divisible by 11
Hence, 7023643 is not divisible by 11.

(e) 58334661
Solution:
58334661
Sum of the digits at odd places is 5 + 3 + 4 + 6 = 18
Sum of the digits at even places is 8 + 3 + 6 + 1 = 18
Difference 18 – 18 = 0
Hence, 58334661 is divisible by 11.

(f) 602111213
Solution:
602111213
Sum of the digits at odd places is 6 + 2+1 + 2 + 3 = 14
Sum of the digits at even places is 0 + 1 + 1 + 1 = 3
Difference 14 – 3 = 11 divisible by 11
Hence, 602111213 is divisible by 11. Question 5.
Fill in the blanks:
(a) A number is divisible by 6 if it is divisible by its two co-prime factors _______ and _______.
Solution:
2, 3

(b) 43185 is divisible by 15 as it is divisible by _______ and _______.
Solution:
3, 5

(c) The number 8625 is not divisible by 6 as it is divisible by _______ but not by _______.
Solution:
3, 2

(d) The number 54420 is divisible by 12 as it is divisible by _______ and _______.
Solution:
4, 3

(e) The number 781022 is divisible by 11 as the difference of the sum of the digits at odd places and the sum of the digits at even places is _______.
Solution:
0 Question 6.
Replace ___ by a digit so that the number is divisible by 9.

(a) 384 ___ 62
Solution:
4

(b) 1 ___ 80498
Solution:
6

(c) 9080 ___
Solution:
1

(d) 46 ___ 21
Solution:
5 Question 7.
Write ‘True’ or ‘False’ for the following statements:

(a) If a number is divisible by 3, it must be divisible by 9.
Solution:
False

(b) If a number is divisible by 18, it must be divisible by 6 and 3.
Solution:
True

(c) If a number is divisible by both 9 and 10, then it must be divisible by 90.
Solution:
True

(d) All numbers which are divisible by 8 are divisible by 4.
Solution:
True

(e) If a number is exactly divisible by two numbers separately then it must be exactly divisible by their sum.
Solution:
False