## RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2

**Other Exercises**

**Question 1.**

Given that the number (overline{35 a 64}) is divisible by 3, where a is a digit, what are the possible volues of a ?**Solution:**

The number (overline{35 a 64}) is divisible by 3

∵The sum of its digits will also be divisible by 3

∴ 3 + 5 + a + b + 4 is divisible by 3

⇒ 18 + a is divisible by 3

⇒ a is divisible by 3 (∵ 18 is divisible by 3)

∴ Values of a can be, 0, 3, 6, 9

**Question 2.**

If x is a digit such that the number (overline { 18×71 }) is divisible by 3,’ find possible values of x.**Solution:**

∵ The number (overline { 18×71 })

is divisible by 3

∴ The sum of its digits will also be divisible by 3

⇒ l + 8+ x + 7 + 1 is divisible by 3

⇒ 17 + x is divisible by 3

The sum greater than 17, can be 18, 21, 24, 27…………

∴ x can be 1, 4, 7 which are divisible by 3.

**Question 3.**

If is a digit of the number (overline { 66784x }) such that it is divisible by 9, find the possible values of x.**Solution:**

∵ The number 66784 x is divisible by 9

∴ The sum of its digits will also be divisible by 9

⇒ 6+6+7+8+4+x is divisible by 9

⇒ 31 + x is divisible by 9

Sum greater than 31, are 36, 45, 54………

which are divisible by 9

∴ Values of x can be 5 on 9

∴ x = 5

**Question 4.**

Given that the number (overline { 67 y 19 }) is divisible by 9, where y is a digit, what are the possible values of y ?**Solution:**

∵ The number (overline { 67 y 19 }) is divisible by 9

∴The sum of its digits will also be divisible by 9

⇒ 6 + 7+ y+ 1+ 9 is divisible by 9

⇒ 23 + y is divisible by 9

∴ The numbers greater than 23 are 27, 36, 45,……..

Which are divisible by 9

∴y = A

**Question 5.**

If (overline { 3 x 2 }) is a multiple of 11, where .v is a digit, what is the value of * ?**Solution:**

∵ The number (overline { 3 x 2 }) is multiple of 11

∴ It is divisible by 11

∴ Difference of the sum of its alternate digits is zero or multiple of 11

∴ Difference of (2 + 3) and * is zero or multiple of 11

⇒ If x – (2 + 3) = 0 ⇒ x-5 = 0

Then x = 5

**Question 6.**

If (overline { 98125 x 2 }) is a number with x as its tens digits such that it is divisible by 4. Find all the possible values of x.**Solution:**

∵ The number (overline { 98125 x 2 }) is divisible by 4

∴ The number formed by tens digit and units digit will also be divisible by 4

∴ (overline { x2 }) is divisible by 4

∴ Possible number can be 12, 32, 52, 72, 92

∴ Value of x will be 1,3, 5, 7, 9

**Question 7.**

If x denotes the digit at hundreds place of the number (overline { 67 x 19 }) such that the

number is divisible by 11. Find all possible values of x.**Solution:**

∵ The number (overline { 67 x 19 }) is divisible by 11

∴ The difference of the sums its alternate digits will be 0 or divisible by 11

∴ Difference of (9 + x + 6) and (1 + 7) is zero or divisible by 11

⇒ 15+x-8 = 0, or multiple of 11,

7 + x = 0 ⇒ x = -7, which is not possible

∴ 7 + x = 11, 7 + x = 22 etc.

⇒ x=11-7 = 4, x = 22 – 7

⇒ x = 15 which is not a digit

∴ x = 4

**Question 8.**

Find the remainder when 981547 is divided by 5. Do this without doing actual division.**Solution:**

A number is divisible by 5 if its units digit is 0 or 5

But in number 981547, units digit is 7

∴ Dividing the number by 5,

Then remainder will be 7 – 5 = 2

**Question 9.**

Find the remainder when 51439786 is divided by 3. Do this without performing actual division.**Solution:**

In the number 51439786, sum of digits is 5 + 1+ 4 + 3 + 9 + 7 + 8 + 6 = 43 and the given number is divided by 3.

∴ The sum of digits must by divisible by 3

∴ Dividing 43 by 3, the remainder will be = 1

Hence remainder = 1

**Question 10.**

Find the remainder, without performing actual division when 798 is divided by 11.**Solution:**

Let n = 798 = a multiple of 11 + [7 + 8 – 9] 798 = a multiple of 11 + 6

∴ Remainder = 6

**Question 11.**

Without performing actual division, find the remainder when 928174653 is divided by 11.**Solution:**

Let n = 928174653

= A multiple of 11+(9 + 8 + 7 + 6 + 3)-(2 + 1+4 + 5)

= A multiple of 11 + 33 – 12

= A multiple of 11 + 21

= A multiple of 11 + 11 + 10

= A multiple of 11 + 10

∴ Remainder =10

**Question 12.**

Given an example of a number which is divisible by :

(i) 2 but not by 4.

(ii) 3 but not by 6.

(iii) 4 but not by 8.

(iv) both 4 and 8 but not 32.**Solution:****(i) 2 but not by 4**

A number is divisible by 2 if units do given is even but it is divisible by 4 if the number formed by tens digit and ones digit is divisible by 4.

∴ The number can be 222, 342 etc.**(ii) 3 but not by 6**

A number is divisible by 3 if the sum of its digits is divisible by 3

But a number is divisible by 6, if it is divided by 2 and 3 both

∴ The numbers can be 333, 201 etc.**(iii) 4 but not by 8**

A number is divisible by 4 if the number formed by the tens digit and ones digit is divisible by 4 but a number is divisible by 8, if the number formed by hundreds digit, tens digit and ones digit is divisible by 8.

∴ The number can be 244, 1356 etc.**(iv) Both 4 and 8 but not by 32**

A number in which the number formed by the hundreds, tens and one’s digit, is divisible by 8 is divisible by 8. It will also divisible by 4 also.

But a number when is divisible by, 4 and 8 both is not necessarily divisible by 32 e.g., 328, 5400 etc.

**Question 13.**

Which of the following statements are true ?

(i) If a number is divisible by 3, it must be divisible by 9.

(ii) If a number is divisible by 9, it must be divisible by 3.

(iii) If a number is divisible by 4, it must be divisible by 8.

(iv) If a number is divisible by 8, it must be divisible by 4.

(v) A number is divisible by 18, if it is divisible by both 3 and 6.

(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.

(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.

(viii) If a number divides three numbers exactly, it must divide their sum exactly.

(ix) If two numbers are co-priirie, at least one of them must be a prime number.

(x) The sum of two consecutive odd numbers is always divisible by 4.**Solution:**

(i) False, it is not necessarily that it must divide by 9.

(ii) Trae.

(iii) False, it is not necessarily that it must divide by 8.

(iv) True.

(v) False, it must be divisible by 9 and 2 both.

(vi) True.

(vii) False, it is not necessarily.

(viii)True.

(ix) False. It is not necessarily.

(x) True.

Hope given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2 are helpful to complete your math homework.

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