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**Playing with Numbers Class 8 Questions And Answers **provided here. These NCERT Solutions are created by expert team at craeers360 keeping in mind of the latest syllabus and pattern of CBSE 2023-23. You have studied various types of numbers such as Integer Numbers, Natural Numbers, Whole Numbers, and Rational Numbers. In NCERT solutions for Class 8 Maths chapter 16 Playing with Numbers, you will find some interesting tricks to find divisibility of numbers. Important topics like divisibility by 10, 5, 2, 9, and 3, numbers in general form, letters for digits are covered in this chapter.

This Story also Contains

- NCERT solutions for Class 8 Maths Chapter 16 Playing with Numbers
- Playing with Numbers Class 8 Questions And Answers PDF Free Download
- Playing with Numbers Class 8 Solutions - Important Formulae
- Playing with Numbers Class 8 NCERT Solutions (Intext Questions and Exercise)
- Playing with numbers class 8 NCERT solutions - Topics
- NCERT Solutions for Class 8 Maths - Chapter Wise
- Key Features Of Playing With Numbers Class 8 NCERT Solutions
- NCERT Solutions for Class 8 - Subject Wise
- Check NCERT Books and NCERT Syllabus

Here you will get the detailed NCERT Solutions for Class 8 Maths by clicking on the link. In Playing with Numbers class 8 solutions, there are some games with numbers and puzzles on the arithmetic problems given in the textbook which makes this chapter very interesting. To strengthen the foundation of the maths you must develop the ability to play with the numbers. For that, you should practice questions given in the textbook. You can take help from NCERT solutions for Class 8 Maths chapter 16 Playing with Numbers that are prepared and explained in a detailed manner.

Divisibility by 2: A number is divisible by 2 when its one's digit is 0, 2, 4, 6, or 8.

Divisibility by 3: A number is divisible by 3 when the sum of its digits is divisible by 3.

Divisibility by 4: A number is divisible by 4 when the number formed by its last two digits is divisible by 4.

Divisibility by 5: A number is divisible by 5 when its one's digit is 0 or 5.

Divisibility by 6: A number is divisible by 6 when it is divisible by both 2 and 3.

Divisibility by 9: A number is divisible by 9 when the sum of its digits is divisible by 9.

Divisibility by 10: A number is divisible by 10 when its one's digit is 0.

Divisibility by 11: A number is divisible by 11 when the difference of the sum of its digits in odd places and the sum of its digits in even places is either 0 or a multiple of 11.

Free download **NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers **for CBSE Exam.

**Class 8 maths chapter 16 NCERT solutions - Topic 16.2 Numbers in General Form **

** Question:****1(i) ** Write the following numbers in generalised form.

25

** Answer: **

any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b

hence,

25 = 10*2 + 5 = 20 + 5

** Question:****1(ii) ** Write the following numbers in generalised form.

73

** Answer: **

any two digit number ab made of digits a and b can be written as ab = 10 × a + b = 10a + b

hence generalized form of the number

73 = 10*7 + 3

** Question:1(iii) ** ** ** Write the following numbers in generalised form.

129

** Answer: **

A 3 digit number madeup of digits a, b, c will be written as,

abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c

hence generalised form of number

129 = 100*1 + 10*2 + 1*9

** Question:****1(iv) ** Write the following numbers in generalised form.

302

** Answer: **

A 3-digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c

hence generalised form of

302 = 100*3 + 10*0 + 1*2

** Question:2(i) ** ** ** Write the following in the usual form.

** Answer: **

As we know ab = 10 × a + b = 10a + b

usual form of number

10*5 + 6 = 50 + 6 = 56

** Question:****2(ii) ** Write the following in the usual form.

** Answer: **

As we know abc = 100 × a + 10 × b + 1 × c

the usual form of number

100*7 + 10*1 + 8 = 700 + 10 + 8 = 718.

** Question:2(iii) ** ** ** Write the following in the usual form.

** Answer: **

As we know abc = 100 × a + 10 × b + 1 × c

casual form of 100 × a + 10 × b + 1 × c = abc.

** NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers - Topic 16.3 Games with Numbers **

** Question:1 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

27

** Answer: **

As we know

(10a + b) + (10b + a) = 11a + 11b = 11 (a + b)

here a = 2 and b = 7

27 + 72 = 99= 11 *9 = a multiple of 11

** Question:2 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

39

** Answer: **

39 + 93 = 133 = 11*12 = a multiple of 11

this can be explained by

(10a + b) + (10b + a) = 11a + 11b = 11 (a + b)

Here a = 3 and b = 9

** Question:3 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

64

** Answer: **

64 + 46 = 110 = 11*10= a multiple of 11

this can be explained by

(10a + b) + (10b + a) = 11a + 11b = 11 (a + b)

here a = 6 and b = 4

** Question:4 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

17

** Answer: **

17 + 71 = 88 = 11*8 = a multiple of 11

this can be explained by

(10a + b) + (10b + a) = 11a + 11b = 11 (a + b)

here a = 1 and b = 7

** Class 8 playing with numbers NCERT solutions - Topic 16.3 Games with Numbers **

** Question:1 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

17

** Answer: **

If the tens digit is larger than the ones digit (that is, a > b), then

(10a + b) – (10b + a) = 10a + b – 10b – a

= 9a – 9b = 9(a – b).

If the unit digit is larger than the tens digit (that is, b > a), he does:

(10b + a) – (10a + b) = 9(b – a).

here a = 1 and b = 7

71 - 17 = 54 = 9*6 = multiple of 9

** Question:2 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

21

** Answer: **

If the tens digit is larger than the ones digit (that is, a > b), then

(10a + b) – (10b + a) = 10a + b – 10b – a

= 9a – 9b = 9(a – b).

If the unit digit is larger than the tens digit (that is, b > a), he does:

(10b + a) – (10a + b) = 9(b – a).

here a = 2 and b = 1

21 - 12 = 9 = 9*1 = multiple of 9

** Question:3 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

96

** Answer: **

If the tens digit is larger than the ones digit (that is, a > b), then

(10a + b) – (10b + a) = 10a + b – 10b – a

= 9a – 9b = 9(a – b).

If the unit digit is larger than the tens digit (that is, b > a), he does:

(10b + a) – (10a + b) = 9(b – a).

here a = 9 and b = 6

96 - 69 = 27= 9*3= multiple of 9

** Question:4 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

** ** 37

** Answer: **

(10a + b) – (10b + a) = 10a + b – 10b – a

= 9a – 9b = 9(a – b).

If the unit digit is larger than the tens digit (that is, b > a), he does:

(10b + a) – (10a + b) = 9(b – a).

here a = 3 and b = 7

73 - 37 = 36 = 9*4 = multiple of 9

** NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers - Topic 16.3 Games with Numbers **

132

** Answer: **

Let's assume the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• If a > c, then the difference between the original number & reversed number

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).

• If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).

• If a & c are equal,then the difference is 0.

here a = 1, b = 3 and c = 2

231 - 132 = 99 = multiple of 99

** ** 469

** Answer: **

Let the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• If a > c, then the difference between the original numbedr & reversed numbers is

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).

• If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).

• if a & c are equal, then the difference is 0.

here a,b and c are 4, 6 & 9 respectively.

964 - 469 = 495 = 99*5 = multiple of 99

** ** 737

** Answer: **

Let the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• If a > c, then the difference between the original number & reversed number is

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).

• If c > a, then the difference between the numbers is (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).

• If A 7 C are equal, the difference is 0.

here a = 7, b = 3 and c = 7

737- 737 = 0= multiple of 99

901

** Answer: **

Let the 3-digit number chosen by Minakshi = 100a + 10b + c.

After reversing the order of the digits, number = 100c + 10b + a.

On subtraction:

• If a > c, then the difference between the original number & reversed number is

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a = 99a – 99c = 99(a – c).

• If a & c are equal, the difference is 0.

here a = 9, b = 0 and c = 1

901- 109= 792= 99*8 = multiple of 99

quotient in each case = c - a.

** NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers - Topic 16.3 Games with Numbers **

** Question:1 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

** ** 417

** Answer: **

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

After adding all the above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c),

It will be divisible by 37 becuase 37 is present in the equation.

here a = 4, b = 1, and c = 7

417 + 741 + 147 = 1332 = 37*36 i.e. divisible by 37.

** Question:2 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

632

** Answer: **

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

The addition of all the above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37

here a = 6, b = 3, and c = 2

632+ 263+ 362= 1221= 37*33 i.e. divisible by 37.

** Question:3 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

117

** Answer: **

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

The addition of the above all three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37

here a = 1, b = 1, and c = 7

117 + 711 + 117 = 999 = 37*27 i.e. divisible by 37.

** Question:4 ** Check what the result would have been if Sundaram had chosen the numbers shown below.

937

** Answer: **

Let choosen number be abc then,

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

The addition of all above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37

here a = 9, b = 3, and c = 7

937 + 793 + 397 = 2109 = 37*57 i.e. divisible by 37.

** NCERT class 8 maths chapter 16 question answer - Exercise: 16.1 **

** Question:1 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

Here we are adding two numbers and unit place of the first number and the second number is A and 5 respectively. unit place of the answer is 2 so the way we can get this result is when we get 12 on adding unit places of both number i.e.

A + 5 = 12

which implies A = 12 - 5 = 7.

Ten's digit of both numbers are 3 and 2.remainder=1

so ten's digit of the answer(B) = 3 + 2 + 1 = 6

Hence A= 7 and B = 6.

** Question:2 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

Here

answer's unit place = 3 , possible addition of unit places digit = 13

A + 8 = 13

A = 13 - 8 = 5

remainder = 1

Ten's place of answer = 4 + 9 + 1 = 14

B = 4

remainder = 1

100's place = C = 1

Hence value of A = 5 , B = 4 and C = 1.

** Question:3 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

Here

first clue :

we have A = a number which when multiplied by itself gives the same number in the unit digit.

possible numbers = 1 and 6

Second Clue:

number when multiplied with 1 and added with the reminder of previous multiplication( A*A) = 9

both first and second clue implies that A = 6.

** Question:4 ** Find the values of the letters in each of the following and give reasons for the steps involved

** Answer: **

There can be two cases

1. when the addition of unit place digit doesn't produce Carry

A + 3 = 6

A = 3

However, to get 3 in unit place of our answer our B has to be 6 and that would produce carry hence this case is not possible.

2. when the addition of unit place digit produces Carry

A + 3 +1 = 6

A = 2

for getting 2 in unit place of answer we need the sum of unit digit of numbers = 12

B + 7 = 12

B = 5

Hence A = 2 and B = 5

** Question:5 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

Here multiplication of 3 and B gives a number whose unit place digit is B .

Possible value of B = 0 and 5

let B = 5

3 * A + 1 = CA

this is not possible for any value of A.

Hence B = 0

now A * 3 = CA ( a number whose unit place digit is A itself when multiplied by 3) hence

possible value of A = 5 and 0

since AB is a two digit number A can not equal to 0.

hence A = 5

A * 3 + 1 = CA

5 * 3 + 1 = 15

hence C = 1

A = 5, B = 5 and C = 1.

** Question:6 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

Here , multiplication of B and 5 gives a number whose ones digit is B. this is possible when B = 0 or 5

let B = 5

B * 5 = 5 * 5 = 25

Carry = 2

5*A + 2 = CA , This os possible only when A = 2 or 7

when A = 2 ,

5*2 + 2 = 12 which implies C = 1

when A = 7

5*7 + 2 = 37 which implies C = 3

now B = 0

B*5 = 0*5 = 0

Carry = 0 so

5 * A = CA which is possible when A = 0 or 5 Howerver A cannot equal to 0 since AB is a two digit number

so A = 5

5*5 = 25 which implies C = 2

hence possible values of A B and C are

A = 5, 2, 7 , B = 0, 5, 5 and C = 2, 1, 3

** Question:7 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

The product of 6 & B gives a number whose unit digit is B again.

possible value of B = 0. 2 , 4, 6 or 8

If B = o then our product will be zero. hence this value of B is not possible.

If B = 2, then B x 6 = 12 . Carry for next step = 1.

6A + 1 = BB = 22

implies A = 21/6 = not any integer value hence this case is also not possible.

If B = 6 then B *6 = 36 and 3 will be carry for next step.

6A + 3 = BB = 66

implies A = 63/6 = not an integer value hence this case is also not possible.

If B = 8 then B * 6 = 48 and 4 will be carry for next step.

6A + 4 = BB = 88

implies A = 14 however A is a single digit number hence this case is also not possible.

If B = 4 then B*6 = 24 and 2 will be carry for next step.

6A + 2 = BB = 22

implies A = 7 .

Hence A = 7 and B = 4 is the correct answer.

** Question:8 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

The addition of 1 and B gives a number whose ones digit is 0. this is possible when digit B = 9 .

1 + B = 10 and 1 is the Carry for the next step

Now, A + 1 + 1 = B => 9

Implies A = 7.

Hence A = 7 and B = 9 is Correct answer

** Question:9 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

The additiion of B and 1 is 8 is giving a number whose ones digit is 8. this means digit B is 7 .

B + 1 = 8 and no carry for next step.

next step :

Now, A + B = 1 => A + 7

which implies A = 4

A + B = 11 and 1 is carry for next step

1 + 2 + A = B

1 + 2 + 4 = 7

Hence A = 4 and B = 7 is correct answer.

** Question:10 ** Find the values of the letters in each of the following and give reasons for the steps involved.

** Answer: **

The addition of A and B is giving a number whose ones digit is 9. The sum can only be 9 not 10 as a sum of two single digits cannot exceed 18. hence there will not be any carry for the next step

2 + A = 0

implies A = 8

2 + 8 = 10 and 1 is the carry for next step.

1 + 1 + 6 = A = 8

it satisfies hence A = 8 and B = 1 is the correct answer.

** NCERT playing with numbers class 8 solutions - Topic 16.5.2 Divisibility By 5 **

** Answer: **

The detailed solution for the above-written problem is as follows,

The unit digit, when divided by 5, must be leaving a remainder of 3. So the unit digit must be either 3 or 8.

** Question:2 ** ** ** If the division leaves a remainder of 1, what might be the one’s digit of ?

** Answer: **

The detailed solution for the above-written question is as follows

If a number is divisible by 5 then it's unit digit must be 0 or 5. so if we need the remainder of 1 when divided by 5 then the numbers unit digit must be 1 or 6.

** Question: ** ** 3 ** If the division leaves a remainder of 4, what might be the one’s digit of N?

** Answer: **

The detailed solution for the above-written question is as follows

If the unit digit of a number is 0 or 5, then it is divisible by 5. hence if we need the remainder of 4 then unit digit of number should be 4 or 9.

** Playing with numbers class 8 NCERT solutions - Topic 16.5.3 Divisibility By 2 **

** Answer: **

The detailed solution for the above-written question is as follows

N is odd; so it's unit digit is odd. Therefore, the unit digit must be 1, 3, 5, 7 or 9.

** Question:2 ** If the division leaves no remainder (i.e., zero remainders), what might be the one’s digit of N?

** Answer: **

The detailed solution for the above-written question is as follows

N is Even; so it's unit digit is even. Therefore, the unit digit must be 2, 4, 6, 8 or 0.

** Answer: **

Since N leaves the remainder of 4 when divided by 5. the possible values in ones place of number N are 4 or 9.

now, since it leaves a remainder of 1 when divided by 2, the N would be an odd number. hence ones digit of N is also an odd number. which means ones digit of our number N is 9.

**NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers - Topic 16.5.4 Divisibility By 9 And 3 **

** Question:1 ** Check the divisibility of the following numbers by 9.

108

** Answer: **

Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9.

Sum of digit of 108 = 1 + 0 + 8 = 9 which is divisible by 9 i.e (9/9 = 1).

hence we conclude 108 is divisible by 9.

** Question:2 ** Check the divisibility of the following numbers by 9.

616

** Answer: **

Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9.

Sum of digits of 616 = 6 + 1 + 6 = 13 which is not divisible by 9,

Hence we conclude 616 is not divisible by by 9.

** Question:3 ** Check the divisibility of the following numbers by 9.

294

** Answer: **

Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9.

Sum of digits of 294 = 2 + 9 + 4 = 15 which is not divisible by 9.

Hence we conclude 294 is not divisible by 9.

** Question:4 ** Check the divisibility of the following numbers by 9.

** ** 432

** Answer: **

Sum of digit 432 = 4 + 3 + 2 = 9 which is divisible by 9.

Hence we conclude 432 is divisible by 9.

** Question:5 ** Check the divisibility of the following numbers by 9.

927

** Answer: **

Sum of digits of 927 = 9 + 2 + 7 = 18 which is divisible by 18.

Hence we conclude number 927 is divisible by 9.

**NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers - Topic 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 * ?

** Answer: **

Yes, it has been prooved that if a number is divisible by any number * m * , then it will also be divisible by each of the factors of * m. *

Let's Assume n is divisible by m, and m is divisible by k.This means

n=pm and

m = qk where all are integers

Now,

n = p(qk) =( pq)k which means n is divisible by k.

Hence 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 as

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

Is it necessary that should be divisible by 11?

** Answer: **

let the number abc be 132

Here a = 1, b = 3 and c = 2

132= 100*1 + 10*3 + 2 = 99 + 11*3 + (1 - 3 + 2)

= 11(9*1+3) + (1 - 3 + 2 )

if number is divisible by 11 then (a - b + c ) must be divisible by 11.

as in above case of number 132 the a - b + c = 1 -3 + 2 = 0 whichis divisible by 11.

Hence we conclude ( a - b + c ) should be divisible by 11 if abc is divisible by 11.

** Question:2(ii) ** Write a 4-digit number as

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

** Answer: **

If the number abcd is divisible by 11 then [ (b + d) - (a + c) ] also must be divisible by 11.

let the number be 1089

here a = 1, b = 0, c = 8 and d = 9

1089 = 1000*1 + 100*0 + 10*8 + 9

= (1001*1 + 99*0 + 11*8) + [(0 + 9) - (1 + 8)]

= 11(91*1 + 9*0 + 8) + [ 9 - 9 ]

here [ (b + d) - (a + c) ] = [9 - 9 ] = 0 which is divisible by 11.

hence If the number abcd is divisible by 11 then [ (b + d) - (a + c) ] also must be divisible by 11.

** Answer: **

Yes,

A number will always 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.

So, for instance, 2728 has the alternating sum of digits 2-7+2-8 = -11. Here -11 is divisible by 11, so is 2728.

Similarly, for 31415, the alternating sum of digits is 3-1+4-1+5 = 10. This would not divisible by 11, so neither is 31415.

** NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers - Topic 16.5.4 Divisibility By 9 And 3 **

** Question:1 ** Check the divisibility of the following numbers by 3.

108

** Answer: **

Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3.

Sum of digits of number 108 = 1 + 0 +8 = 9 which is divisible by 3.

Hence we conclude number 108 is divisible by 3.

** Question:2 ** Check the divisibility of the following numbers by 3.

616

** Answer: **

Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3.

Sum of digits of number 616 = 6 + 1 + 6 = 13 which is not divisible by 3.

Hence we conclude number 616 is not divisible by 3.

** Question:3 ** Check the divisibility of the following numbers by 3.

294

** Answer: **

Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3.

Sum of digits of number 294= 2 + 9 + 4 = 15 which is divisible by 3.

Hence we conclude number 294 is divisible by 3.

** Question:4 ** Check the divisibility of the following numbers by 3.

432

** Answer: **

Sum of digits of number 432 = 4 + 3 + 2 = 9 which is divisible by 3.

Hence we conclude number 432 is divisible by 3.

** Question:5 ** Check the divisibility of the following numbers by 3.

927

** Answer: **

Sum of digits of number 927 = 9 + 2 + 7 = 18 which is divisible by 3.

Hence we conclude number 108 is divisible by 3.

**NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers - Exercise: 16.2 **

** Question:1 ** If is a multiple of 9, where y is a digit, what is the value of ?

** Answer: **

If a number is a multiple of 9 , then the sum of its digit will be divisible by 9.

Sum of digits of 21y5= 2 + 1 + y + 5 = 8 + y

8 + y is a multiple of 9 when y = 1 (since y can only be single digit )

hence y = 1 is correct answer.

** Answer: **

If a number is a multiple of 9, then the sum of its digit will be divisible by 9 .

Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z

hence 9 + z have to be multiple of 9

this possible when 9 + z = 0, 9 , 18, 27 ...

since z is a single digit number this sum can only be 9 or 18 . therefore, z should be 0 or 9 .

hence two possible values of z = 0 or 9.

** Question:3 ** If is a multiple of 3, where is a digit, what is the value of ?

(Since is a multiple of 3, its sum of digits is a multiple of 3; so is one of these numbers: 0, 3, 6, 9, 12, 15, 18, ... . But since is a digit, it can only be that . Therefore, . Thus, can have any of four different values.)

** Answer: **

if a number is multiple of 3 then the sum of its digits is also a multiple of 3.

Summ of digits of 24x = 2 + 4 + x = 6 + x

6 + x is a multiple of 3 which means

6 + x = 0 or 3 or 9 or 12....

but since x is a single digit the possible sum is 6 or 9 or 12 or 15 and hence value of x are 0 or 3 or 6 or 9 respectively.

hence possible value of x = 0, 3, 6, or 9.

** Question:4 ** If is a multiple of 3, where is a digit, what might be the values of ?

** Answer: **

If a number is multiple of 3, the sum of its digits will be multiple of 3.

Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z

9 + z is a multiple of 3 and since z is a single digit

9 + z is any one of 9, 12, 15 or 18

thus value of z is 0, 3, 6 or 9 respectively.

Hence the possible value of z is 0,3,6 or 9.

- Numbers in General Form
- Games with Numbers
- Letters for Digits
- Tests of Divisibility

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Chapter -2 | |

Chapter-3 | |

Chapter-4 | |

Chapter-5 | |

Chapter-6 | |

Chapter-7 | |

Chapter-8 | |

Chapter-9 | |

Chapter-10 | |

Chapter-11 | |

Chapter-12 | |

Chapter-13 | |

Chapter-14 | |

Chapter-15 | |

Chapter-16 | Playing with Numbers |

**Comprehensive Coverage:** Maths chapter 16 class 8 solutions cover all topics and concepts related to playing with numbers as per the Class 8 syllabus.

**Step-by-Step Solutions:** Class 8 maths ch 16 question answer are detailed, step-by-step explanations for each problem, making it easy for students to understand and apply mathematical concepts related to number patterns, divisibility, and prime factors.

**Variety of Problems:** A wide range of problems, including exercises and additional questions, to help students practice and test their understanding of number patterns and properties.

1. What are the important topics of ch 16 maths class 8 Playing with Numbers ?

Numbers in general form, letters for digits, and divisibility test are the important topics of this chapter.

2. Does CBSE class maths is tough ?

In CBSE class 8 you will study basic and very simple maths where most of the topics related to the previous classes.

3. How many chapters are there in the CBSE class 8 maths ?

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

4. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

5. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

6. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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