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**Question 1.** If the division N Ã· 5 leaves a remainder of 3, what might be the one,s digit of N?**Solution:** The oneâ€™s digit, when divided by 5, must leave a remainder of 3. So the oneâ€™s digit must be either 3 or 8.**Question 2.** If the division N Ã· 5 leaves a remainder of 1, what might be the oneâ€™s digit of N?**Solution:** If remainder = 1, then the one,s digit of â€˜Nâ€™ must be either 1 or 6.**Question 3.** If the division N Ã· 5 leaves a remainder of 4, what might be the oneâ€™s digit of N?**Solution:** If remainder = 4, then the one,s digit of â€˜Nâ€™ must be either 9 or 4.**DIVISIBILITY BY 2**

If oneâ€™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 Ã· 2 leaves a remainder of 1, what might be the oneâ€™s digit of N?**Solution:** N is odd; so its oneâ€™s digit is odd. Therefore, the oneâ€™s digit must be 1, 3, 5, 7 or 9.**Question 2. **If the division N Ã· 2 leaves no remainder (i.e, zero remainder), what might be the oneâ€™s digit of N?**Solution:**

âˆµ Remainder = 0

âˆ´ Oneâ€™s digit can be 0, 2, 4, 6 or 8.**Question 3. **Suppose that the division NÃ· 5 leaves a remainder of 4 and the division N Ã· 2 leaves a remainder of 1. What must be the oneâ€™s digit of N?**Solution: **

âˆµ N Ã· 5 and remainder = 4

âˆ´ Oneâ€™s digit can be 4 or 9.

Again N Ã· 2 and remainder = 1

âˆ´ N must be an odd number.

Thus, oneâ€™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

âˆµ 1 + 0 + 8 = 9

and 9 is divisible by 9. [âˆµ 9 Ã· 9 = 1 and remainder = 0]

âˆ´ 108 is divisible by 9.**2.** 616

We have 6 + 1 + 6 = 13

and 13 Ã· 9 = 1, remainder = 4

i.e., 13 is not divisible 9.

âˆ´ 616 is also not divisible by 9.**3**. 294

We have 2 + 9 + 4 = 15

and 15 Ã· 9 = 1, remainder = 6

i.e. 15 is not divisible by 9.

âˆ´ 294 is also not divisible by 9.**4. **432

We have 4 + 3 + 2 = 9

9 Ã· 9 = 1, remainder = 0

âˆ´ 432 is divisible by 9.**5.** 927

We have 9 + 2 + 7 = 18

and 18 Ã· 9 = 2, remainder = 0

i.e. 18 is divisible by 9.

âˆ´ 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 â€˜mâ€™.**Question 2.**

(i) Write a 3-digit number abc as 100a + 10b + c

= 99a + 11b + (a â€“ b + c)

= 11(9a + b) + (a â€“ b + c)

If the number abc is divisible by 11, then what can you say about (a â€“ b + c)?

Is it necessary that (a + c â€“ b) should be divisible by 11?

(ii) Write a 4-digit number abcd as 1000a + 100b + 10c + d

= (1001a + 99b + 11c) â€“ (a â€“ b + c â€“ d)

= 11(91a + 9b + c) + [(b + d) â€“ (a + c)]

If the number abcd is divisible by 11, then what can you say about

[(b + d) â€“ (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 â€“ b + c) showed be divisible by 11.

(ii) (b + d) â€“ (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 Ã· 3 = 6, remainder = 0

âˆ´ 108 is divisible by 3. [âˆµ 18 is divisible by 3]**2.** 616

We have 6 + 1 + 6 = 13

and 13 Ã· 3 = 4, remainder = 1

âˆ´ 13 is not divisible by 3.

Thus 616 is also not divisible by 3.**3.** 294

We have 2 + 9 + 4 = 15

and 15 Ã· 3 = 5, remainder = 0

âˆ´ 15 is divisible by 3.

Thus, 294 is also divisible by 3.**4.** 432

We have 4 + 3 + 2 = 9

and 9 Ã· 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 Ã· 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:** âˆµ We have 2 + 1 + y + 5 = 8 + y

21y5 is a multiple of 9,

âˆ´ (8 + y) must be divisible by 9.

âˆ´ (8 + y) should be 0, 9, 18, 27, ..., etc.

âˆ´ 8 + y = 0 is not required.

Since, y is a digit

8 + y = 9 fi y = 9 â€“ 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

âˆµ 31z5 is divisible by 9.

âˆ´ (9 + z) must be equal to 0, or 9 or 18 or 27, â€¦

But z is a digit.

âˆ´ 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

âˆµ 31z5 is divisible by 3.

âˆ´ (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.

âˆ´ If 9 + z = 0, then z = â€“9.

If 9 + z = 3, then z = â€“6.

If 9 + z = 6, then z = â€“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.

âˆ´ Possible values of z are 0, 3, 6, or 9.

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