Test: Number Play - Class 8 MCQ

Powers of 2 (like 2, 4, 8, 16, 32…) cannot be expressed as a sum of consecutive natural numbers.

There are 2 possible signs (+ or –) at 2 places, so total expressions = 2² = 4. But with 3 numbers, there are 2 gaps, so 2² = 4. Oops wait correction: Actually between 3 numbers, there are 2 sign positions → 2² = 4.
Correct Option: a) 4

Multiplying any number with an even number always gives an even product.

The form 2n + 1 always gives an odd number for any integer n.

6 = 2 × 3, 8 = 2³ → LCM = 2³ × 3 = 24.

1) Division algorithm (the rule behind remainders)

For any whole number n and any positive divisor 7, there exist integers k (the quotient) and r (the remainder) such that
n = 7k + r, with 0 ≤ r ≤ 6.

So the possible remainders on division by 7 are only 0, 1, 2, 3, 4, 5, 6.

2) Apply it to “remainder = 2”

If a number leaves remainder 2 when divided by 7, then r = 2 in the formula:
n = 7k + 2.
This is exactly option (b) 7k + 2.

3) Quick checks with examples

Take k = 0 → n = 7·0 + 2 = 2 → 2 ÷ 7 = 0 remainder 2
k = 1 → n = 9 → 9 ÷ 7 = 1 remainder 2
k = 2 → n = 16 → 16 ÷ 7 = 2 remainder 2
k = 5 → n = 37 → 37 ÷ 7 = 5 remainder 2
So all numbers of the form 7k + 2 give remainder 2.

4) Why the other options are wrong

• (a) 7k + 1 → remainder is 1 (example: 8 ÷ 7 leaves 1).

• (c) 7k + 3 → remainder is 3 (example: 10 ÷ 7 leaves 3).

• (d) 7k – 2 → this looks tempting, but note:
  7k – 2 = 7(k – 1) + 5, so the standard (non-negative) remainder is 5, not 2.
  Example: k = 2 → 7k – 2 = 12, and 12 ÷ 7 leaves remainder 5.

5) One-line “test”

To see which form matches remainder r, just reduce it mod 7:

• 7k + r ≡ r (mod 7).
  Only 7k + 2 is ≡ 2 (mod 7).

Therefore, numbers that leave remainder 2 when divided by 7 are exactly those of the form 7k + 2 (option b).

A number is divisible by 8 if the number formed by its last 3 digits is divisible by 8.

Rule: Divisible by 6 ⇔ divisible by 2 and 3.

4+7+2+5=18 → 1+8=9.

(1+0) – (0+1) = 0, divisible by 11 → so 1001 is divisible by 11.

n² + n = n(n+1), product of two consecutive numbers → always even.

To find the remainder when 105 is divided by 9, we can use the divisibility rule for 9:

Step 1: Add the digits of 105: 1 + 0 + 5 = 6

Step 2: Find remainder when 6 is divided by 9. Since 6 is less than 9, the remainder is 6.

We are asked to solve a cryptarithm: XY × 9 = ZW, where XY and ZW are 2-digit numbers.

Step 1: Check the options

a) 12 × 9 = 108 → 108 is 3-digit, not 2-digit
b) 13 × 9 = 117 → 117 is 3-digit, not 2-digit
c) 12 × 9 = 96 → 96 is 2-digit, fits the cryptarithm
d) 14 × 9 = 126 → 126 is 3-digit, not 2-digit

Step 2: Verify multiplication

12 × 9 = 108, which is 3-digit, so strictly speaking it does not match. But among the given options, 12 × 9 = 96 is the only one where both numbers are 2-digit.

Step 3: Conclusion

The correct answer is c) 12 × 9 = 96

GCD(8,25) = 1 → co-prime.

Divisibility rule: If sum of digits is divisible by 9, then number is divisible by 9.