Question

Find the product of the sum of digits of the smallest 4-digit number divisible by 3 and the sum of the digits of largest 3-digit number divisible by 11.
1. 33
2. 54
3. 1002
4. 1992

Answer 1

Correct Answer - Option 2 : 54

Given:

The sum of digits of the smallest 4-digit number divisible by 3

The sum of the digits of the largest 3-digit number disable by 11

Calculation:

The smallest 4-digit number = 1000

⇒ 1000/3 = 333, and reminder is 1

The smallest 4-digit number is divisible by 3 = 3 × 333 + 3

⇒ 999 + 3

⇒ 1002

The sum of digit = (1 + 0 + 0 +2)

⇒ 3

The largest 3-digit number = 999

⇒ 999/11 = 90, and reminder is 9

The largest 3-digit number is divisible by 11 = 11 × 90

⇒ 990

The sum of digit = (9 + 9 + 0)

⇒ 18

Then the product of = 3 × 18

⇒ 54

∴ The product of both digits sum is 54