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NCERT Solutions for Class 8 Maths - Playing with Numbers - 2

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.

The document NCERT Solutions for Class 8 Maths - Playing with Numbers - 2 is a part of the Class 8 Course NCERT Textbooks & Solutions for Class 8.
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FAQs on NCERT Solutions for Class 8 Maths - Playing with Numbers - 2

1. What are the different types of numbers that can be played with?
Ans. Playing with numbers involves various types of numbers such as prime numbers, composite numbers, even numbers, odd numbers, etc.
2. Can you explain the concept of prime numbers?
Ans. Prime numbers are numbers that are greater than 1 and can only be divided by 1 and themselves without leaving any remainder. For example, 2, 3, 5, 7, etc., are prime numbers.
3. How do you determine if a number is even or odd?
Ans. A number is even if it is divisible by 2 without leaving a remainder. On the other hand, a number is odd if it is not divisible by 2 and leaves a remainder.
4. What is the significance of composite numbers in playing with numbers?
Ans. Composite numbers are numbers that have more than two factors. These numbers are important in playing with numbers as they can be further divided into prime factors.
5. Can you provide an example of playing with numbers?
Ans. Sure! Let's consider the number 12. It is a composite number as it has factors other than 1 and itself (1, 2, 3, 4, 6, and 12). We can further break it down into its prime factors, which are 2, 2, and 3.
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