Class 8 Mathematics – Number Play | Exam-Ready Practice Worksheet
Topics Covered
- Consecutive Numbers
- Parity: Odd and Even Numbers
- Factors and Multiples
- Always, Sometimes or Never
- Remainders
- Divisibility by 2, 3, 4, 5, 6, 8, 9, 10 and 11
- Digital Roots
- Cryptarithms / Digits in Disguise
Section A – MCQs (10 × 1 = 10 Marks)
Q1. The sum of four consecutive numbers is 34. The numbers are:
Q2. Which expression is always even for every integer value of n?
Q3. Which number leaves remainder 3 when divided by 5?
Q4. If a number is divisible by both 9 and 4, then it must be divisible by:
Q5. Which number is divisible by 9?
Q6. A number is divisible by 11 when the difference between the sums of alternate digits is:
Q7. The digital root of 5832 is:
Q8. In a cryptarithm, the same letter represents:
Q9. In a cryptarithm, the first digit of a number:
Q10. If a divides M and a divides N, then a also divides:
Section B – Very Short Answer (8 × 2 = 16 Marks)
17, 22, 27, 32, ______, ______, ______
n, n + 1, n + 2, n + 3. If their sum is 54, find the four numbers.
Q13. Without actually calculating, determine whether each expression is even or odd:
- 8m + 6
- 7p + 4
- 12q − 8
- 5r + 7
Q14. Find the remainder when:
- 47 is divided by 5
- 83 is divided by 7
- 125 is divided by 6
Q17. Find the digital root of:
- 48,729
- 99,999
- 12,345
Section C – Short Answer Questions (6 × 3 = 18 Marks)
The sum of five consecutive numbers is 125. Find all five numbers and verify your answer.
Four consecutive integers are represented by n, n + 1, n + 2, n + 3. Prove that the sum of these four numbers is always even.
Q21. Always, Sometimes or Never: Determine whether each statement is Always True, Sometimes True or Never True. Give an example or counterexample.
- The sum of two even numbers is divisible by 4.
- The sum of two multiples of 4 is a multiple of 4.
- If a number is divisible by 7, it is divisible by every multiple of 7.
Q22. Remainders:
A number leaves remainder 3 when divided by 7. Another number leaves remainder 5 when divided by 7. Without finding the actual numbers, determine the remainder when their:
- sum is divided by 7
- difference is divided by 7
Explain your reasoning.
Find all possible values of x so that 52×4 is divisible by 9. Show the reasoning using the divisibility rule.
Q24. Divisibility by 11:
Determine whether each number is divisible by 11. If it is not, find the remainder using the divisibility shortcut.
- 841
- 5529
- 90,904
Section D – HOTS / Reasoning
Section E – Cryptarithms / Digits in Disguise
Rules:
- Each letter represents a digit from 0 to 9.
- The same letter represents the same digit throughout.
- Different letters represent different digits.
- The first digit of a number cannot be 0.
Q27. Solve the cryptarithm:
+ 1B
—-
B0
Find the values of A and B.
Q28. Solve:
+ 37
—-
6A
Find A and B.
Q29. Solve:
× 8
—-
RS
Find P, Q, R and S.
Q30. Solve:
× 5
—-
BC
Find A, B and C.
Q31. Challenge: Solve:
× 6
—-
KKK
Find J and K.
Bonus Challenge
Q32. Solve the cryptarithm:
× E
—-
GGG
Find all possible values of E, F and G.
Exam Tips
- For parity questions, represent even and odd numbers algebraically.
- For remainder questions, work with the known remainders instead of calculating the original numbers.
- For divisibility by 3 or 9, use the sum of the digits.
- For divisibility by 11, use the alternating-sum method.
- For cryptarithms, start from the units column and track carries carefully.
- In a cryptarithm, different letters must represent different digits and a leading digit cannot be zero.
- For Always/Sometimes/Never questions, use algebra and counterexamples where appropriate.
