Important Questions Class 8 Maths Chapter 5 Number Play

Important Questions Class 8 Maths Chapter 5 Number Play

Number Play is Chapter 5 of the Class 8 Maths book Ganita Prakash. It explores puzzles and patterns with numbers. Students study even and odd numbers, divisibility rules, number tricks and simple algebraic reasoning about numbers.

This page gives 12 important questions from Class 8 Maths Chapter 5 Number Play with step-by-step answers prepared by Extramarks experts. The questions cover even and odd numbers, divisibility tests, remainders, cryptarithms and reasoning with expressions. Solve each question on paper first and then check your working with the answer.

Q1. During homework, a student notices that some algebraic expressions always give even answers while others change between even and odd depending on the value chosen. The student decides to analyse expressions using structure instead of substituting values repeatedly.

(a) The expression 5n + 3 gives an odd result for n = 2. Determine whether this will remain odd for all values of n and justify your answer.

(b) Decide whether the expression n² + n can ever be odd and explain your reasoning.

Answer: (a) If n is even, then 5n is even and 5n + 3 is odd.

If n is odd, then 5n is odd and 5n + 3 is even.

So the expression does not remain odd for all values of n.

(b) n² + n = n(n + 1)

One of n or n + 1 must be even since they are consecutive numbers.

Hence, their product is always even and never odd.

Q2. Explain why the product of two odd numbers is always odd.

Answer: Let the two odd numbers be 2a + 1 and 2b + 1.

So, the product of the two odd numbers = (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1

This can be written as 2(2ab + a + b) + 1.

Since the result is not divisible by 2, the product is always odd.

Q3. Determine the parity of the sum of an even number and an odd number. Justify your answer.

Answer: Let the even number be 2k and the odd number be 2m + 1.

Their sum is 2(k + m) + 1, which is not divisible by 2.

Hence, the sum of an even number and an odd number is odd.

Q4. Solve the cryptarithm given below, where A and B are digits.

Answer: AB can be written as 10A + B and 6A can be written as 60 + A.

So, 10A + B + 37 = 60 + A ⇒ 9A + B = 23

If A = 3, then 27 + B = 23 gives B = −4, which is not possible.

If A = 2, then 18 + B = 23 ⇒ B = 5, which is possible.

So, A = 2 and B = 5.

Q5. If the number 59a4b is divisible by 18, list all possible pairs of values of a and b.

Answer: For a number to be divisible by 18, it must be divisible by both 2 and 9.

Divisibility by 2: the last digit b must be even. So, b can be 0, 2, 4, 6 or 8.

Divisibility by 9: sum of digits = 5 + 9 + a + 4 + b = 18 + a + b. Since 18 is already a multiple of 9, a + b must also be a multiple of 9.

If b = 0, then a = 9; if b = 2, then a = 7; if b = 4, then a = 5; if b = 6, then a = 3; if b = 8, then a = 1.

Hence, the possible pairs (a, b) are: (9, 0), (7, 2), (5, 4), (3, 6), (1, 8)

Q6. A number leaves remainder 6 when divided by 7. Find the remainder when

(a) the number is multiplied by 3

(b) 5 is subtracted from the number

Answer: Let the number be 7k + 6.

(a) 3(7k + 6) = 21k + 18 = 21k + 14 + 4 = 7(3k + 2) + 4. So, the remainder is 4.

(b) 7k + 6 − 5 = 7k + 1. So, the remainder is 1.

Q7. If a 4-digit number leaves a remainder of 5 when divided by 9, what remainder will the number formed by reversing its digits leave when divided by 9? Justify your answer.

Answer: Reversing digits does not change the sum of digits, and the remainder on division by 9 depends only on the sum of digits.

So, the remainder when divided by 9 remains the same.

Hence, the reversed number also leaves remainder 5 when divided by 9.

Q8. A class is making number cards for a board game. Certain moves are allowed only if a statement about divisibility is always true. To avoid memorising, students must decide whether each claim is Always True, Sometimes True, or Never True using reasoning and counterexamples.

(a) Decide whether the statement “The sum of two multiples of 6 is always a multiple of 6” is Always True, Sometimes True, or Never True. Justify.

(b) Decide whether the statement “The product of a multiple of 6 and a multiple of 3 is always a multiple of 9” is Always True, Sometimes True, or Never True. Justify your answer.

Answer: (a) Always True.

Let the two multiples of 6 be 6a and 6b. Their sum = 6a + 6b = 6(a + b), which is again a multiple of 6. The reasoning does not depend on the particular values, so it holds always.

(b) Always True.

Let the numbers be 6a and 3b. Their product = 18ab = 9(2ab), which is a multiple of 9.

Q9. In the school carnival, entry passes are printed with a 3-digit code N. For a quick sorting trick, the volunteer checks two rules without doing long division. First, the code is grouped into one of 7 counters based on its remainder when divided by 7. Second, a “special lane” is used if the code is divisible by 11. You are given that N leaves remainder 3 when divided by 7.

(a) Without calculating N, determine the remainder when (N + 28) is divided by 7.

(b) Without calculating N, determine the remainder when (4N − 5) is divided by 7.

Answer: (a) Since N leaves remainder 3 when divided by 7, write N = 7k + 3.

N + 28 = 7k + 3 + 28 = 7(k + 4) + 3. So, (N + 28) leaves remainder 3 when divided by 7.

(b) 4N − 5 = 4(7k + 3) − 5 = 28k + 7 = 7(4k + 1), which is exactly divisible by 7.

So, (4N − 5) leaves remainder 0 when divided by 7.

Q10. Check whether 342 is divisible by 6. Justify your answer using divisibility rules.

Answer: To be divisible by 6, a number must be divisible by both 2 and 3.

342 is even, so it is divisible by 2.

The sum of its digits is 3 + 4 + 2 = 9, which is divisible by 3.

Hence, 342 is divisible by 6.

Q11. Explain why a number divisible by 12 must be divisible by both 3 and 4.

Answer: The number 12 can be written as 3 × 4. A number divisible by 12 is a multiple of 12, so it is also a multiple of 3 and of 4. Divisibility by 3 depends on the digit sum and divisibility by 4 depends on the last two digits. Only when both conditions are satisfied is the number divisible by 12.

Q12. Explain why a number divisible by 6 must be divisible by both 2 and 3.

Answer: The number 6 can be written as 2 × 3. A number divisible by 6 is a multiple of 6, so it is also a multiple of 2 and of 3. Divisibility by 2 means the number is even, and divisibility by 3 means the sum of its digits is divisible by 3.

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