Number Play — Class 8 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 8 Mathematics chapter "Number Play" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 8 Mathematics chapter "Number Play" — 8 important questions with detailed answers for CBSE board exam prep…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 8 Mathematics — Number Play. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Number Play — All 8 Questions Solved
Q1. Check if 7845 is divisible by 3 without performing the division.
We use the divisibility rule for 3.
Step 1: Recall the divisibility rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3.
Step 2: Find the sum of digits of 7845
Digits: 7, 8, 4, 5
Sum = 7 + 8 + 4 + 5 = 24
Step 3: Check if 24 is divisible by 3
24 ÷ 3 = 8 (no remainder)
✓ 24 is divisible by 3
Step 4: Conclusion
Since the sum of digits (24) is divisible by 3, the number 7845 is divisible by 3.
Step 5: Verification (optional)
7845 ÷ 3 = 2615
Answer: Yes, 7845 is divisible by 3 because the sum of its digits (24) is divisible by 3
Q2. Write the first five terms of the pattern: 2, 4, 8, 16, ...
We need to identify the pattern and write five terms.
Step 1: Analyze the sequence
Term 1: 2
Term 2: 4
Term 3: 8
Term 4: 16
Step 2: Find the relationship between consecutive terms
4 = 2 × 2
8 = 4 × 2
16 = 8 × 2
Each term is 2 times the previous term.
Step 3: This is a geometric sequence with first term a = 2 and common ratio r = 2
General term: aₙ = 2 × 2ⁿ⁻¹ = 2ⁿ
Step 4: Write the first five terms
Term 1: 2¹ = 2
Term 2: 2² = 4
Term 3: 2³ = 8
Term 4: 2⁴ = 16
Term 5: 2⁵ = 32
Step 5: Verify
Each term is double the previous term ✓
Answer: The first five terms are: 2, 4, 8, 16, 32
Q3. Check if 5830 is divisible by 10.
We use the divisibility rule for 10.
Step 1: Recall the divisibility rule for 10
A number is divisible by 10 if its units digit (last digit) is 0.
Step 2: Check the last digit of 5830
The number is 5830
Last digit = 0
Step 3: Conclusion
Since the last digit is 0, the number 5830 is divisible by 10.
Step 4: Verification
5830 ÷ 10 = 583 (no remainder)
Answer: Yes, 5830 is divisible by 10 because its units digit is 0
Q4. Find the pattern and write the next three terms: 3, 6, 12, 24, ...
We need to find the rule for this sequence and extend it.
Step 1: Analyze the sequence
Term 1: 3
Term 2: 6
Term 3: 12
Term 4: 24
Step 2: Find the relationship between consecutive terms
6 = 3 × 2
12 = 6 × 2
24 = 12 × 2
Each term is 2 times the previous term.
Step 3: This is a geometric sequence with first term a = 3 and common ratio r = 2
General term: aₙ = 3 × 2ⁿ⁻¹
Step 4: Find the next three terms
Term 5: 24 × 2 = 48
Term 6: 48 × 2 = 96
Term 7: 96 × 2 = 192
Or using the formula:
Term 5: 3 × 2⁴ = 3 × 16 = 48
Term 6: 3 × 2⁵ = 3 × 32 = 96
Term 7: 3 × 2⁶ = 3 × 64 = 192
Answer: The next three terms are: 48, 96, 192
Q5. Is 9201 divisible by 9? Use the divisibility rule.
We use the divisibility rule for 9.
Step 1: Recall the divisibility rule for 9
A number is divisible by 9 if the sum of its digits is divisible by 9.
Step 2: Find the sum of digits of 9201
Digits: 9, 2, 0, 1
Sum = 9 + 2 + 0 + 1 = 12
Step 3: Check if 12 is divisible by 9
12 ÷ 9 = 1 remainder 3
✗ 12 is NOT divisible by 9
Step 4: Conclusion
Since the sum of digits (12) is not divisible by 9, the number 9201 is not divisible by 9.
Step 5: Verification
9201 ÷ 9 = 1022.333... (not a whole number)
Answer: No, 9201 is not divisible by 9 because the sum of its digits (12) is not divisible by 9
Q6. Write the pattern rule for: 1, 4, 9, 16, 25, ... in algebraic form.
We need to identify what the pattern represents and write its rule.
Step 1: Analyze the sequence
Term 1: 1
Term 2: 4
Term 3: 9
Term 4: 16
Term 5: 25
Step 2: Recognize these as perfect squares
1 = 1²
4 = 2²
9 = 3²
16 = 4²
25 = 5²
Step 3: Write the pattern
For the nth term, the value is n²
Step 4: General formula
aₙ = n²
where n = position of the term (1st, 2nd, 3rd, ...)
Step 5: Verify
a₁ = 1² = 1 ✓
a₂ = 2² = 4 ✓
a₃ = 3² = 9 ✓
a₄ = 4² = 16 ✓
a₅ = 5² = 25 ✓
Answer: The pattern rule is aₙ = n² (the square of the position number)
Q7. Check the divisibility of 4328 by 4 using the divisibility rule.
We use the divisibility rule for 4.
Step 1: Recall the divisibility rule for 4
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
Step 2: Identify the last two digits of 4328
The number is 4328
Last two digits: 28
Step 3: Check if 28 is divisible by 4
28 ÷ 4 = 7 (no remainder)
✓ 28 is divisible by 4
Step 4: Conclusion
Since the last two digits (28) form a number divisible by 4, the number 4328 is divisible by 4.
Step 5: Verification
4328 ÷ 4 = 1082 (no remainder)
Answer: Yes, 4328 is divisible by 4 because its last two digits (28) form a number divisible by 4
Q8. Find the missing number: 2, 5, 10, 17, 26, ?, 50
We need to find the pattern and identify the missing term.
Step 1: Analyze the differences between consecutive terms
Term 1: 2
Term 2: 5 (difference: 5 - 2 = 3)
Term 3: 10 (difference: 10 - 5 = 5)
Term 4: 17 (difference: 17 - 10 = 7)
Term 5: 26 (difference: 26 - 17 = 9)
Term 6: ?
Term 7: 50
Step 2: Analyze the differences
First differences: 3, 5, 7, 9, ?, ?
These are odd numbers in sequence!
Step 3: Continue the pattern of differences
Next difference = 11
Next difference = 13
Step 4: Find the missing term (Term 6)
Term 6 = Term 5 + 11 = 26 + 11 = 37
Step 5: Verify
Term 7 = Term 6 + 13 = 37 + 13 = 50 ✓
Step 6: Recognize the formula
aₙ = n² + 1
a₁ = 1² + 1 = 2 ✓
a₂ = 2² + 1 = 5 ✓
a₃ = 3² + 1 = 10 ✓
a₄ = 4² + 1 = 17 ✓
a₅ = 5² + 1 = 26 ✓
a₆ = 6² + 1 = 37 ✓
a₇ = 7² + 1 = 50 ✓
Answer: The missing number is 37
Number Play — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following numbers is divisible by 5?
- A. 123
- B. 457
- C. 680
- D. 901
Answer: C. 680 — A number is divisible by 5 if its unit digit is 0 or 5. In 680, the unit digit is 0.
MCQ 2. The general form of a two-digit number 'xy' where x is the tens digit and y is the units digit is:
- A. x + y
- B. 10x + y
- C. x * y
- D. 10y + x
Answer: B. 10x + y — In a two-digit number 'xy', x represents 10 times its value and y represents its unit value, so it is 10x + y.
MCQ 3. A number is divisible by 2 if its unit digit is:
- A. 1
- B. 3
- C. 5
- D. An even number
Answer: D. An even number — The divisibility rule for 2 states that a number is divisible by 2 if its unit digit is 0, 2, 4, 6, or 8, which are all even numbers.
MCQ 4. Which digit should replace '*' in the number 23* to make it divisible by 10?
- A. 0
- B. 2
- C. 5
- D. None of these
Answer: A. 0 — A number is divisible by 10 if its unit digit is 0. Replacing '*' with 0 makes the number 230.
MCQ 5. If the sum of the digits of a number is divisible by 3, then the number itself is divisible by:
- A. 2
- B. 5
- C. 3
- D. 9
Answer: C. 3 — According to the divisibility rule for 3, if the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
MCQ 6. The general form of a three-digit number 'abc' where a, b, c are digits is:
- A. a + b + c
- B. 10a + 10b + c
- C. 100a + 10b + c
- D. a * b * c
Answer: C. 100a + 10b + c — In a three-digit number 'abc', 'a' is in the hundreds place, 'b' in tens, and 'c' in units, so it's 100a + 10b + c.
MCQ 7. If a number is divisible by both 2 and 3, then it is also divisible by:
- A. 4
- B. 5
- C. 6
- D. 8
Answer: C. 6 — If a number is divisible by two co-prime numbers (like 2 and 3), it is divisible by their product, which is 6.
MCQ 8. Which of the following is always true about the difference between a two-digit number (10a+b) and the number obtained by reversing its digits (10b+a)?
- A. It is always divisible by 2
- B. It is always divisible by 5
- C. It is always divisible by 9
- D. It is always divisible by 11
Answer: C. It is always divisible by 9 — The difference is (10a+b) - (10b+a) = 9a - 9b = 9(a-b), which is always divisible by 9.
MCQ 9. If the number 5x2 is divisible by 3, what is the possible value of x?
- A. 0
- B. 1
- C. 2
- D. 5
Answer: C. 2 — For 5x2 to be divisible by 3, the sum of its digits (5 + x + 2) must be divisible by 3. So, 7 + x must be divisible by 3. If x=2, then 7+2=9, which is divisible by 3.
MCQ 10. If the number 7A8 is divisible by 4, what is the possible value of A?
- A. 0
- B. 1
- C. 3
- D. 5
Answer: A. 0 — A number is divisible by 4 if the number formed by its last two digits is divisible by 4. So, 'A8' must be divisible by 4. If A=0, 08 (which is 8) is divisible by 4.
MCQ 11. In the addition problem: A B + 3 7 = 6 A, find the digit A.
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — From the units column, B + 7 results in A with a carry. From the tens column, A + 3 + (carry from units) = 6. If there is no carry from units, A+3=6 means A=3, then B+7=3, impossible. So, a carry of 1 occurs from units. Thus, A + 3 + 1 = 6, which gives A = 2.
MCQ 12. The sum of a two-digit number (10a+b) and the number obtained by reversing its digits (10b+a) is always divisible by:
- A. 2
- B. 7
- C. 9
- D. 11
Answer: D. 11 — The sum is (10a+b) + (10b+a) = 11a + 11b = 11(a+b), which is always divisible by 11.
MCQ 13. If 1x5 is a three-digit number divisible by 9, what is the value of x?
- A. 1
- B. 2
- C. 3
- D. 5
Answer: C. 3 — For 1x5 to be divisible by 9, the sum of its digits (1 + x + 5) must be divisible by 9. So, 6 + x must be divisible by 9. If x=3, then 6+3=9, which is divisible by 9.
MCQ 14. Which digit should replace 'B' in the number 1B0 to make it divisible by 8?
- A. 1
- B. 2
- C. 3
- D. 5
Answer: B. 2 — A number is divisible by 8 if the number formed by its last three digits is divisible by 8. So, 1B0 must be divisible by 8. We check options: If B=2, then 120 is divisible by 8 (120 ÷ 8 = 15).
MCQ 15. If the number 2Y3 is divisible by 11, what is the digit Y?
- A. 0
- B. 1
- C. 2
- D. 5
Answer: D. 5 — For a number to be divisible by 11, the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) must be 0 or a multiple of 11. So, (3 + 2) - Y = 5 - Y must be a multiple of 11. For Y=5, 5-5=0, which is a multiple of 11.
MCQ 16. In the multiplication problem: A x 3 = B A, where B A is a two-digit number. What is the digit A?
- A. 0
- B. 1
- C. 5
- D. 7
Answer: C. 5 — From the units column, A x 3 must end in A. Possible digits are A=0 or A=5. If A=0, B A would be 0, not a two-digit number. So A must be 5. Then 5 x 3 = 15, so B=1 and A=5.
MCQ 17. What is the remainder when the number 5346 is divided by 9?
- A. 0
- B. 1
- C. 3
- D. 6
Answer: A. 0 — To find the remainder when a number is divided by 9, we find the remainder of the sum of its digits divided by 9. Sum of digits = 5 + 3 + 4 + 6 = 18. Since 18 is divisible by 9, the remainder is 0.
MCQ 18. If the number 12Y4 is divisible by 4, what is the smallest possible non-zero value of Y?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — For 12Y4 to be divisible by 4, the number formed by its last two digits, 'Y4', must be divisible by 4. We check the options: If Y=1, 14 is not divisible by 4. If Y=2, 24 is divisible by 4. Thus, the smallest non-zero value for Y is 2.
MCQ 19. Consider a three-digit number 'abc'. If (100a + 10b + c) - (a + b + c) is always divisible by:
- A. 9
- B. 10
- C. 11
- D. 100
Answer: A. 9 — (100a + 10b + c) - (a + b + c) = 99a + 9b = 9(11a + b). This expression is always divisible by 9.
MCQ 20. In the subtraction puzzle: A B - 2 5 = 4 6, find the digit B.
- A. 1
- B. 2
- C. 3
- D. 4
Answer: A. 1 — From the units column, to get 6 from B-5, we must have borrowed from A. So (10 + B) - 5 = 6, which implies B + 5 = 6, so B=1. (A - 1) - 2 = 4, so A - 3 = 4, A=7. Thus B=1.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.