NCERT Solutions (Part- 2)- Playing with Numbers Class 8 Notes | EduRev

Class 8 Mathematics by Full Circle

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Class 8 : NCERT Solutions (Part- 2)- Playing with Numbers Class 8 Notes | EduRev

The document NCERT Solutions (Part- 2)- Playing with Numbers Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by Full Circle.
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Question 1. If the division N ÷ 5 leaves a remainder of 3, what might be the one,s digit of N?
Solution: The one’s digit, when divided by 5, must leave a remainder of 3. So the one’s digit must be either 3 or 8.

Question 2. If the division N ÷ 5 leaves a remainder of 1, what might be the one’s digit of N?
Solution: If remainder = 1, then the one,s digit of ‘N’ must be either 1 or 6.

Question 3. If the division N ÷ 5 leaves a remainder of 4, what might be the one’s digit of N?
Solution: If remainder = 4, then the one,s digit of ‘N’ must be either 9 or 4.

DIVISIBILITY BY 2
If one’s digit of a number is 0, 2, 4, 6 or 8, then the number is divisible by 2.

Question 1. If the division N ÷ 2 leaves a remainder of 1, what might be the one’s digit of N?
Solution: N is odd; so its one’s digit is odd. Therefore, the one’s digit must be 1, 3, 5, 7 or 9.

Question 2. If the division N ÷ 2 leaves no remainder (i.e, zero remainder), what might be the one’s digit of N?
Solution:
∵ Remainder = 0
∴ One’s digit can be 0, 2, 4, 6 or 8.

Question 3. Suppose that the division N÷ 5 leaves a remainder of 4 and the division N ÷ 2 leaves a remainder of 1. What must be the one’s digit of N?
Solution: 
∵ N ÷ 5 and remainder = 4
∴ One’s digit can be 4 or 9.
Again N ÷ 2 and remainder = 1
∴ N must be an odd number.
Thus, one’s digit can be 9 only.

DIVISIBILITY BY 9 AND 3
(i) A number N is divisible by 9 if the sum of its digits is divisible by 9, otherwise it is not divisible by 9.
(ii) A number N is divisible by 3, if the sum of its digits is divisible by 3, otherwise it is not divisible by 9.
Note: A number divisible by 9 is also divisible by 3.

Question: Check the divisibility of the following number by 9.
1. 108
2. 616
3. 294
4. 432
5. 927

Solution: 
1. 108
     ∵                1 + 0 + 8 = 9
and 9 is divisible by 9.                         [∵ 9 ÷ 9 = 1 and remainder = 0]
∴ 108 is divisible by 9.

2. 616
We have   6 + 1 + 6 = 13
and               13 ÷ 9 = 1, remainder = 4
i.e., 13 is not divisible 9.
∴ 616 is also not divisible by 9.

3. 294
We have         2 + 9 + 4 = 15
and                15 ÷ 9 = 1, remainder = 6
i.e. 15 is not divisible by 9.
∴ 294 is also not divisible by 9.

4. 432
We have          4 + 3 + 2 = 9
                        9 ÷ 9 = 1, remainder = 0
∴ 432 is divisible by 9.

5. 927
We have          9 + 2 + 7 = 18
and                18 ÷ 9 = 2, remainder = 0
i.e. 18 is divisible by 9.
∴ 927 is also divisible by 9.

THINK, DISCUSS AND WRITE 
Question 1. You have seen that a number 450 is divisible by 10. It is also divisible by 2 and 5 which are factors of 10. Similarly, a number 135 is divisible 9. It is also divisible by 3 which is a factor of 9. Can you say that if a number is divisible by any number m, then it will also be divisible by each of the factors of m?
Solution: Yes, if a number is divisible by any number m, then it will also be divisible by each of the factors of ‘m’.

Question 2.
(i) Write a 3-digit number abc as 100a + 10b + c
                                                                           = 99a + 11b + (a – b + c)
                                                                           = 11(9a + b) + (a – b + c)
If the number abc is divisible by 11, then what can you say about (a – b + c)?
Is it necessary that (a + c – b) should be divisible by 11?

(ii) Write a 4-digit number abcd as 1000a + 100b + 10c + d
                                                   = (1001a + 99b + 11c) – (a – b + c – d)
                                                   = 11(91a + 9b + c) + [(b + d) – (a + c)]
If the number abcd is divisible by 11, then what can you say about
[(b + d) – (a + c)]?

(iii) From (i) and (ii) above, can you say that a number will be divisible by 11 if the different between the sum of digits at its odd places and that of digits at the even
places is divisible by 11?

Solution:
(i) Yes, it is necessary that (a – b + c) showed be divisible by 11.
(ii) (b + d) – (a + c) is divisible by 11.
(iii) Yes, a number will be divisible by 11 if the difference between the sum of digits at
its odd places and that of digits at the even places is divisible by 11.


Question: Check the divisibility of the following numbers by 3.
1. 108
2. 616
3. 294
4. 432
5. 927

Solution: 
1. 108
   We have         1 + 0 + 8 = 18
and                         18 ÷ 3 = 6, remainder = 0
∴ 108 is divisible by 3.                            [∵ 18 is divisible by 3]

2. 616
We have          6 + 1 + 6 = 13
and                        13 ÷ 3 = 4, remainder = 1
∴ 13 is not divisible by 3.
Thus 616 is also not divisible by 3.

3. 294
We have           2 + 9 + 4 = 15
and                       15 ÷ 3 = 5, remainder = 0
∴ 15 is divisible by 3.
Thus, 294 is also divisible by 3.

4. 432
We have           4 + 3 + 2 = 9
and                         9 ÷ 3 = 3, remainder = 0
i.e. 9 is divisible by 3.
Thus, 432 is also divisible by 3.

5. 927
We have     9 + 2 + 7 = 18
and                 18 ÷ 3 = 6, remainder = 0
i.e. 18 is divisible by 3.
Thus, 927 is also divisible by 3.


EXERCISE 16.2 
Question 1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution: ∵ We have 2 + 1 + y + 5 = 8 + y
    21y5 is a multiple of 9,
∴ (8 + y) must be divisible by 9.
∴ (8 + y) should be 0, 9, 18, 27, ..., etc.
∴ 8 + y = 0 is not required.
  Since, y is a digit
   8 + y = 9 fi y = 9 – 8 or y = 1

Question 2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Solution: We have 3 + 1 + z + 5 = 9 + z
∵ 31z5 is divisible by 9.
∴ (9 + z) must be equal to 0, or 9 or 18 or 27, …
   But z is a digit.
∴  9 + z = 0 or 9 + z = 18
If 9 + z = 0, then z = 0 and if 9 + z = 18, then z = 9.

Question 3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
Solution: Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, ... But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.

Question 4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution: We have 3 + 1 + z + 5 = 9 + z
∵ 31z5 is divisible by 3.
∴ (9 + z) must be divisible by 3.
i.e. (9 + z) = 0 or 3 or 6 or 9 or 15 or 18
since, z is a digit.
∴ If 9 + z = 0, then z = –9.
   If 9 + z = 3, then z = –6.
   If 9 + z = 6, then z = –3.
   If 9 + z = 9, then z = 0.
   If 9 + z = 12, then z = 3.
   If 9 + z = 15, then z = 6.
   If 9 + z = 18, then z = 9.
   If 9 + z = 21, then z = 12.
∴ Possible values of z are 0, 3, 6, or 9.

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