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NCERT Solutions for Class 8 Maths Chapter 5 Number Play - Exercise 5.4

Mastering complex numerical puzzles is simple with our NCERT Solutions for Class 8 Maths Ch 5: Playing with Numbers – Exercise 5.4. This exercise takes the basic rules of arithmetic and applies them to multi-step multiplication puzzles where multiple letters (A, B, C) are used. You will learn how to use logical elimination and unit-digit properties to solve "Cryptarithms."

Class 8 Math Lessons (Ch 5): Learn to Solve Complex Digit Puzzles with Ease! The solutions to Exercise 5.4 will be presented in a step-by-step approach to help students decode multiplication grids. These solutions are in accordance with the CBSE Guidelines to improve analytical reasoning. Additionally, these solutions build the foundation for decoding and logic-based problem solving in competitive exams like the IMO or NTSE.

1.0Download NCERT Solutions Class 8 Maths Chapter 5 Number Play Ex 5.4 : Free PDF

Our easy-to-follow NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.4 are available for download as a PDF. Perfect for offline study and quick reference.

NCERT Solutions Class 8 Maths Chapter 5 Ex 5.4

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2.0 Detailed NCERT Class 8 Maths Chapter 5 Solutions of Exercise 5.4

1.If 31 z 5 is a multiple of 9 , where z is a digit, what is the value of z ? Explain why there are two answers to this problem.

Sol. Here, 31z5 Sum of the digits $= 3 + 1 + z + 5 = 9 + z$ $(9 + z)$ should be divisible by 9 . $z = 0, 3105$ is divisible by 9 . $z = 9, 3195$ is also divisible by 9 . $∴ z = 0$ or 9 There are two answers to this problem because, excluding z , the sum of the digits is divisible by 9 .

2. "I take a number that leaves a remainder of 8 when divided by 12.1 take another number, which is 4 short of a multiple of 12. Their sum will always be a multiple of 8", claims Snehal. Examine his claim and justify your conclusion.

Sol. A number that leaves a remainder of 8 . when divided by $12 : 12 k + 8$ , where $k \geq 1$ . Also, another number 4 short of a multiple of 12: $12 k - 4$

3. When is the sum of two multiples of 3 , a multiple of 6 , and when is it not? Explain the different possible cases, and generalise the pattern.

Sol. Multiples of 3 are: $3, 6, 9, 12, 15, 18, \dots\dots\dots .$ . $3 + 6 = 9$ , not a multiple of 6 . $6 + 9 = 15$ , not a multiple of 6 . $3 + 9 = 12$ , multiple of 6 . $6 + 12 = 18$ , multiple of 6 . There are two possible cases.

If both numbers are odd, then the sum is a multiple of 6 .
If both numbers are even, then the sum is a multiple of 6 .

4.Sreelatha says, "I have a number that is divisible by 9 . If I reverse its digits, it will still be divisible by $9^{''}$ . (i) Examine if her conjecture is true for any multiple of 9. (ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9 ?

Sol. We know that if the sum of the digits of a number is divisible by 9 , then the numbers is also divisible by 9 . If we reverse its digits, the sum of its digits will not change because the digits are the same. So, the reverse is also divisible by 9 . (i) Let the number be 81 .

So, $81 \to 8 + 1 = 9$ , the number is divisible by 9 . Reverse of 81 is 18 . So, $18 \to 1 + 8 = 9$ , the number is divisible by 9 . (ii) Yes, any other digit shuffles is possible as the number is still a multiple of 9. Let take a number 459 So, $459 \to (4 + 5 + 9) \to 18$ , the number is divisible by 9 . Reverse of 459 is 954. So, $954 \to 9 + 5 + 4 = 18$ , the number is still divisible by 9 .

5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b .

Sol. Given by question, 48a23b is a multiple of 18 . As we know that, If the number is a multiple of 18 , then it is also a multiple of 2 and 9. ∴ 48a23b Sum of the digits $= 4 + 8 + a + 2 + 3 + b = 17 + a + b$

Case 1: Put $a = 1$ and $b = 0$ 481230, it is possible values of a and b . Sum = 18, it is divisible by 9 . Case 2: Put $a = 4$ and $b = 6$ 484236 Sum $= 17 + 10 = 27$ , it is divisible by 9 . Thus, the possible values of a and 6 are a $= 1$ and $b = 0, a = 4$ and $b = 6$ ; there are two possible cases.

6. If $3 p 7 q 8$ is divisible by 44 , list all possible pairs of values for p and q .

Sol. Given by question, 3p7q8 is divisible by 44.

As we know, if a number is divisible by 44 , then it is also divisible by 4 and 11 . ∴ 3p7q8 Case 1: Put $p = 1$ and $q = 0$ 37708 is divisible by 4 and 11, then it is also divisible by 44 . Case 2: Put $p = 5$ and $q = 2$ 35728 is divisible by 4 and 11, then it is also divisible by 44 . Case 3: Put $p = 3$ and $q = 4$ 33748 is divisible by 4 and 11 , then it is also divisible by 44 . Case 4: Put $p = 1$ and $q = 6$ 31768 is divisible by 4 and 11 , then it is also divisible by 11 . Thus, $(p = 7, q = 0), (p = 5, q = 2), (p = 3, q$ = 4), and ( $p = 1$ and $q = 6$ ) are the possible pairs of values for p and q .

7. Find three consecutive numbers such that the first number is a multiple of 2 , the second number is a multiple of 3 , and the third number is a multiple of 4 . Are there more such numbers? How often do they occur?

Sol. Let $x, x + 1$ and $(x + 2)$ be the three numbers Put $x = 2, \Rightarrow 2, 3, 4$ Put $x = 14, \Rightarrow 14, 15, 6$ Put $x = 26, \Rightarrow 26, 27, 28$ Put $x = 38, \Rightarrow 38, 39, 40$ Thus, the three consecutive numbers are ( $14, 15, 16$ ), Put $x = 26, \Rightarrow 26, 27, 28$ $(26, 27, 28)$ and $(38, 39, 40)$ There are infinite numbers, spaced apart by 12 .

8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.

Sol. We know that if a number is a multiple of 36 , then it is also a multiple of 4 and 9 . 45000 Last two digits $= 00$ , it is divisible by 4 . Sum of the digits $= 4 + 5 + 0 + 0 + 0 = 9$ , it is also divisible by 9 . Thus, 45000 is completely divisible by 36 . The five multiples of 36 between 45,000 and 47,000. $(45, 000 + 36), (45, 000 + 2 \times 36), (45, 000 + 3 \times 36), (45, 000 + 4 \times 36)$ and $(45, 000 + 5 \times 36$ ) i.e., $45, 036, 45, 072, 45, 108, 45, 144$ , and 45,180.

9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.

Sol. Given the middle number in the sequence of 5 consecutive even numbers 5p. The other four numbers in the sequence in terms of $p$ are $5 p - 4, 5 p - 2, 5 p + 2$ , $5 p + 4$ Hence, the other four numbers in sequence are p, 3p, 7p, and 9p.

10. Write a 6 -digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6 .

Sol. We know that if the number is divisible by 3 and 5, then it is also divisible by 15 .

Consider the number 643215. Sum of the digits $= 6 + 4 + 3 + 2 + 1 + 5 =$ 21 , which is divisible by 3 .

Thus, 643215 is divisible by 3 . One's place = 5, it is also divisible by 5 . Hence, 643215 is divisible by 15 . One's place is not 0 , because the digits are reversed, it becomes a 5 -digit number.

Lakhs place is always taken as an even number.

Reversed the digits: 512346 One's place $= 6, 512346$ is divisible by 2 . Sum of the digits $= 5 + 1 + 2 + 3 + 4 + 6 =$ 21.

It is also divisible by 3 . Hence, 512346 is divisible by 6 .

11. Deepak claims,"There are some multiples of 11 which, when doubled, are still multiples of 11 . But other multiples of 11 don't remain multiples of 11 when doubled". Examine if his conjecture is true; explain your conclusion.

Sol. The multiples of 11 are: $11, 22, 33, 44$ , 55,...

When doubled, $22, 44, 66, 88, 110, \dots\dots$ i.e. $11 \times 2, 11 \times 4, 11 \times 6, 11 \times 8, 11 \times 10, \dots$ . are also multiples of 11.

False, if multiples of 11 are doubled, then the multiples of 11 are these numbers.

12. Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning. (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9 . (ii) The sum of three consecutive even numbers will be divisible by 6 . (iii) If abcdef is a multiple of 6 , then badcef will be a multiple of 6 . (iv) $8 (7 b - 3) - 4 (11 b + 1)$ is a multiple of 12.

Sol. (i) Always True, The multiple of 6 can be written as 6a, where a is an integer.

The multiple of 3 can be written as 3 b , where b is an integer. $∴$ Product $= (6 a) \times (3 b) = 18 (ab)$ is a multiple of 9. (ii) Always True,

The sum of three consecutive even numbers will be divisible by 6 .

For example $2 + 4 + 6 = 12, 4 + 6 + 8 = 18, 6 + 8 + 10 = 24, 8 + 10 + 12 =$ 30,...

These numbers are divisible by 6 . (iii) Always True, because one's place does not change. (iv) Sometimes true, Conclusion: $8 (7 \times 1 - 3) - 4 (11 \times 1 + 1) = - 16$ , not divisible by 12 . $8 (7 \times 10 - 3) - 4 (4 \times 10 + 1) = 536 - 164 = 372$ , divisible by 12 .

13. Choose any 3 numbers. When is their sum divisible by 3 ? Explore all possible cases and generalise.

Sol. Let the three numbers be $n_{1}, n_{2}$ , and $n_{3}$ . Let their remainders when divided by 3 be $r_{1}, r_{2}$ , and $r_{3}$ .

The sum $n_{1} + n_{2} + n_{3}$ is divisible by 3 if and only if $r_{1} + r_{2} + r_{3}$ is divisible by 3 .

Case 1: All remainders are 0. $r_{1} = 0, r_{2} = 0, r_{3} = 0$ Sum of remainders $= 0 + 0 + 0 = 0$ , which is divisible by 3 .

Case 2: All remainders are 1. $r_{1} = 1, r_{2} = 1, r_{3} = 1$ Sum of remainders $= 1 + 1 + 1 = 3$ , which is divisible by 3 .

Case 3: All remainders are 2. $r_{1} = 2, r_{2} = 2, r_{3} = 2$ Sum of remainders $= 2 + 2 + 2 = 6$ , which is divisible by 3 . Case 4: One remainder is 0 , one is 1 , and one is 2 . $r_{1} = 0, r_{2} = 1, r_{3} = 2$ (in any order). Sum of remainders $= 0 + 1 + 2 = 3$ , which is divisible by 3 .

The sum of three numbers is divisible by 3 if and only if all three numbers have the same remainder when divided by 3 , or if they all have different remainders when divided by 3 .

14. Is the product of two consecutive integers always a multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

Sol. Yes, the product of two consecutive integers is always a multiple of 2 . $1 \times 2 = 2, 2 \times 3 = 6, 5 \times 6 = 30, 10 \times 11 =$ 110, and so on. Since we know that multiplying by an odd number and an even number is always an even number. No, it is not always a multiple of 6 . $1 \times 2 = 2, 4 \times 5 = 20, 7 \times 8 = 56$ Since it is not divisible by 6 . The product of 4 consecutive integers $2 \times 3 \times 4 \times 5 = 120$ , $4 \times 5 \times 6 \times 7 = 840$ , $5 \times 6 \times 7 \times 8 = 1680$ We can say that the product of 4 consecutive integers, divisible by 12 . The product of five consecutive integers is: $1 \times 2 \times 3 \times 4 \times 5 = 120$ , $2 \times 3 \times 4 \times 5 \times 6 = 720$ , $3 \times 4 \times 5 \times 6 \times 7 = 2520$ Hence, we can say that the product of five consecutive integers is always divisible by 24 .

15. Solve the cryptarithms (i) $EF \times E = GGG$ (ii) $WOW \times 5 = MEOW$

Sol. (i) This means a 2 -digit number multiplied by 5 gives a 3 -digit number. 2-digit number = 20, 21,...., 99 $37 \times 3 = 111$ , all conditions are satisfied. (ii) This means a 3-digit number multiplied by 5 gives 4 -digit numbers. Pick 3-digit number $= 200, 201, \dots$ . , 999 $525 \times 5 = 2625$

16. Which of the following Venn diagrams captures the relationship between the multiples of 4,8 , and 32 ? (i)

(ii)

(iii)

(iv)

Sol. (iv) Multiples of 4 are: 4, 8, 12, 16, 20, 24, $28, 32, 36, 40, 44, 48, 52, 56, 60, 64, \dots$ Multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64,....

Multiples of 32 are: $32, 64, 96, 128, \dots$ The Venn diagram captures the relationship between the multiples of 4,8 , and 32 :

3.0 4.0Key Concepts of Chapter 5 Exercise 5.4

Multiplication Cryptarithms: Solving puzzles where A x B = CA or similar patterns.
Unique Digit Rule: In any single puzzle, a specific letter must represent the same digit throughout the calculation.
Carry-over Logic in Multiplication: Understanding how the tens-place digit from a product affects the next column's addition.
Number Patterns: Exploring the properties of numbers like "repunits" (numbers consisting only of the digit 1, like 11, 111, 1111).

5.0NCERT Solutions for Class 8 Maths Chapter 5 Number Play : All Exercises

NCERT Solutions for Class 8 Maths Chapter 5 Number Play - Exercise 5.1

NCERT Solutions for Class 8 Maths Chapter 5 Number Play - Exercise 5.2

NCERT Solutions for Class 8 Maths Chapter 5 Number Play - Exercise 5.3

NCERT Solutions for Class 8 Maths Chapter 5 Number Play - Exercise 5.4

6.0Benefits of NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.4

Conceptual Clarity: Demonstrates how to logically narrow down 10 possible digits to the correct one.
Strategic Thinking: Teaches students to start with the "constraint" (the digit that repeats the most).
Accuracy: Shows the importance of the final verification step to ensure no rules were broken.

NCERT Class 8 Maths Ch. 5 Number Play Other Exercises:-

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 - A Square and a Cube

Chapter 2 - Power play

Chapter 3 - A story of Numbers

Chapter 4 - Quadrilaterals

Chapter 5 - Number Play

Chapter 6 - We distibute, yet thinmgs multiply

Chapter 7 - Proportional Reasoning

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