Worksheet Solutions: Number Play | Worksheets with Solutions for Class 8 PDF Download

• a) 15
• b) 12
• c) 8
• d) 9
• Solution: c) 8 (Powers of 2 like 8 cannot be expressed as sums of consecutive natural numbers, as per the chapter.)

• a) 4
• b) 6
• c) 8
• d) 16
• Solution: c) 8 (There are 3 positions for signs, each can be + or -, so 2^3 = 8 expressions.)

• a) Odd
• b) Even
• c) Depends on the numbers
• d) Always multiple of 3
• Solution: b) Even (Odd ± Odd = Even, as per the parity rules in the chapter.)

• a) 3g + 5h
• b) b²
• c) x + 1
• d) 2a + 2b
• Solution: d) 2a + 2b (Both terms are multiples of 2, so always even; other options can be odd depending on values.)

• a) 12
• b) 18
• c) 36
• d) 72
• Solution: c) 36 (LCM of 9 and 4 is 36, as stated in the chapter.)

• a) 5k + 2
• b) 5k + 3
• c) 5k - 1
• d) 5k
• Solution: b) 5k + 3 (As explained, numbers like 3, 8, 13 are 5k + 3.)

• a) Units digit only
• b) Last two digits
• c) Sum of all digits
• d) Last three digits
• Solution: b) Last two digits (The last two digits must form a number divisible by 4.)

• a) The sum of its digits is divisible by 11
• b) The difference between the sum of odd and even positioned digits is divisible by 11
• c) The last digit is 0 or 5
• d) The number is even
• Solution: b) The difference between the sum of odd and even positioned digits is divisible by 11 (As per the divisibility rule for 11 in the chapter.)

• a) 0
• b) 3
• c) 6
• d) 9
• Solution: d) 9 (For non-zero multiples of 9, the digital root is 9.)

• a) 10 × 8 = 80
• b) 11 × 8 = 88
• c) 12 × 8 = 96
• d) 13 × 8 = 104
• Solution: c) 12 × 8 = 96 (It fits the rules: 2-digit, distinct digits, and product is 2-digit.)

• True/False
• Solution: False (Not all can; e.g., powers of 2 like 2, 4, 8 cannot.)

• True/False
• Solution: False (Powers of 2 cannot be expressed this way.)

• True/False
• Solution: False (Not all; e.g., 2 cannot.)

• True/False
• Solution: True (As given in the chapter: -1 + 0 + 1 = 0.)

• True/False
• Solution: True (All have the same parity, either all even or all odd; for consecutive, they are even.)

• True/False
• Solution: False (It changes by an even number, like 2b.)

• True/False
• Solution: False (Can be odd or even; e.g., g=2, h=1: 6+5=11, odd.)

• True/False
• Solution: True (Sum of multiples of 8 is a multiple of 8.)

• True/False
• Solution: False (Not always; e.g., 30 + 10 = 40, 40 div by 8, but 30 and 10 not.)

• True/False
• Solution: True (Divisibility rule for 3.)

• True/False
• Solution: False (Never; sum is odd, multiples of 6 are even.)

• True/False
• Solution: True (Divisibility rule for 5.)

• True/False
• Solution: True (4+8+9+7+1+0=29, 2+9=11, 1+1=2.)

• True/False
• Solution: False (First digit ≠ 0.)

• True/False
• Solution: True (All div by 9 are div by 3; reverse not true, e.g., 15 div by 3 not 9.)

• Solution: multiple (As stated in the chapter.)

• Solution: even (All results are even numbers.)

• Solution: Even (Even × Anything = Even.)

• Solution: 2m + n (2(2m + n).)

• Solution: 24 (LCM of 6 and 4 is 24.)

• Solution: 3 (5k + 3.)

• Solution: three (Last three digits.)

• Solution: sum (Sum of digits.)

• Solution: 9 (3, 6, 9 cycle.)

• Solution: odd (Sum odd positions - sum even positions.)

• Solution: 3 (Div by 2 and 3.)

• Solution: 1 (10^n ≡ 1 mod 9.)

• Solution: digital (Digital root.)

• Solution: one (Each letter = one digit.)

• Solution: 8 (3 and 8, as 24=3×8 and co-prime.)

1. Find a set of three consecutive natural numbers whose sum is 18.

Solution: Let the numbers be n, n+1, n+2. Their sum is n + (n+1) + (n+2) = 3n + 3 = 18. Solving, 3n = 15, so n = 5. The numbers are 5, 6, 7.

2. Using four consecutive numbers starting from 5 (i.e., 5, 6, 7, 8), calculate the value of the expression 5 + 6 - 7 - 8.

Solution: Compute 5 + 6 - 7 - 8 = 11 - 7 - 8 = 4 - 8 = -4.

3. Evaluate the expression 43 + 37 and determine if the result is even or odd.

Solution: 43 + 37 = 80. Since 80 is divisible by 2, it is even.

4. For integers m = 3 and n = -2, compute the value of 4m + 2n.

Solution: 4m + 2n = 4(3) + 2(-2) = 12 - 4 = 8.

5. Find the least common multiple (LCM) of 9 and 4.

Solution: Prime factors: 9 = 3², 4 = 2². LCM = 2² × 3² = 4 × 9 = 36.

6. Find the smallest positive number of the form 5k + 3 that is greater than 20.

Solution: Solve 5k + 3 > 20. Then, 5k > 17, k > 3.4. Since k is a whole number, try k = 4: 5(4) + 3 = 20 + 3 = 23.

7. Determine the number formed by the last two digits of 5678 and check if it is divisible by 4.

Solution: Last two digits: 78. Check: 78 ÷ 4 = 19.5, not an integer, so not divisible by 4.

8. For the number 462, calculate the difference between the sum of digits in odd and even positions to check divisibility by 11.

Solution: Digits: 4, 6, 2. Odd positions (1st, 3rd): 4 + 2 = 6. Even position (2nd): 6. Difference: 6 - 6 = 0. Since 0 is divisible by 11, 462 is divisible by 11.

9. Calculate the digital root of 729.

Solution: Sum digits: 7 + 2 + 9 = 18. Sum again: 1 + 8 = 9. Digital root is 9.

10. In the cryptarithm PQ × 8 = RS, where PQ and RS are two-digit numbers with distinct digits, compute the product for PQ = 12.

Solution: PQ = 12, so 12 × 8 = 96. Thus, RS = 96.

11. Find the sum of the digits of 99009 and check if it is divisible by 9.

Solution: Digits: 9 + 9 + 0 + 0 + 9 = 27. Check: 27 ÷ 9 = 3, so divisible by 9.

12. Compute the remainder when 427 is divided by 9 using the sum of its digits.

Solution: Digits: 4 + 2 + 7 = 13. Sum again: 1 + 3 = 4. Remainder is 4.

13. For the number 186, check if it is divisible by 6 by testing divisibility by 2 and 3.

Solution: Div by 2: 186 ends in 6 (even), so divisible. Div by 3: 1 + 8 + 6 = 15, 15 ÷ 3 = 5, so divisible. Thus, 186 is divisible by 6.

14. If two numbers, 24 and 40, are both divisible by 8, compute their difference and check if it is divisible by 8.

Solution: Difference: 40 - 24 = 16. Check: 16 ÷ 8 = 2, so divisible by 8.

15. Find the digital root of the 6th multiple of 3.

Solution: 6th multiple: 3 × 6 = 18. Sum digits: 1 + 8 = 9. Digital root is 9.