NCERT solutions for Class 8 Maths Chapter 16 Playing with Numbers

Updated on Aug 3, 2020 - 5:16 p.m. IST by Ravindra Pindel

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers- Maths is all about the numbers and you can simply play with the numbers. You have studied various types of numbers such as integer numbers, natural numbers, whole numbers, and rational numbers. You must have also studied interesting properties about them. 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. In NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers, 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. If you are facing problem while solving exercises of NCERT 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. You can get NCERT solutions from class 6 to 12 by clicking the below link. Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

Let’s take an example of games with numbers- two friends Meenakshi and Sundaram play a game Minakshi asks Sundaram to think a two-digit number and then-

Topics of NCERT Class 8 Maths Chapter 16 Playing with Numbers-

16.1 Introduction
16.2 Numbers in General Form
16.3 Games with Numbers
16.4 Letters for Digits
16.5 Tests of Divisibility

NCERT solutions for class 8 maths chapter 16 playing with numbers for all exercises

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.2 numbers in general form

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

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.

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.

$10\times 5 + 6$

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.

$100\times 7 + 10\times 1 + 8$

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.

$100\times a + 10\times c + b$

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.

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.

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.

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.

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

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.

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.

Answer:

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.

Answer:

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.

Answer:

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

Question:1 Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.

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

Question:2 Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.

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

Question:3 Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.

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

Question:4 Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end.

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 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, 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 solutions for class 8 maths chapter 16 playing with numbers-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 solutions for class 8 maths chapter 16 playing with numbers topic 16.5.2 divisibility by 5

Question:1 If the division N ÷ 5 leaves a remainder of 3, what might be the ones digit of N? (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.)

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 $N\div 5$ leaves a remainder of 1, what might be the one’s digit of $N$ ?

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 $N\div 5$ 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.

NCERT solutions for class 8 maths chapter 16 playing with numbers topic 16.5.3 divisibility by 2

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

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 $N \div 2$ 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.

Question: 3 Suppose that the division $N\div 5$ leaves a remainder of 4, and the division $N\div 2$ leaves a remainder of 1. What must be the one’s digit of N?

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: