Question
A six-digit number 4c68d2 is divisible by 12. Find the
sum of all possible values of 'c' for the largest possible value of 'd'.Solution
ATQ, A number is divisible by 12 when it is divisible by 3 and 4 both. A number is divisible by 4 when its last two digits are divisible by 4. Here 'd2' is divisible by 4 for 'd' = 0, 2, 4, 6, and 8. Largest possible value of 'd' = 8. A number is divisible by 3 when the sum of its digits is divisible by 3. Sum of digits of 4c6882 = 4 + c + 6 + 8 + 8 + 2 = 28 + c. So, possible values of 'c' = 2, 5, and 8. Therefore, the required sum = 2 + 5 + 8 = 15.
(2310.23 ÷ 32.98) + (1008.32 ÷ 23.9) + 1594.11 = ?
? × 32.91 – 847.95 ÷ √16.4 – 13.982 = √24.7 × 24.04
119.98% of 80.02 - 15.12 × 2.02 + 19.95 = ?
25.09 × (√15 + 19.83) = ? of 19.87 ÷ 4.03Â
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
68.98 × 41.03 – (12.33)² + 15.78% of 8398.87 = ? – 40.22
? % of 759.96 + 932.99 = 1237.01
 (3/5) of 3025 + (18² + 12²) = ? + 22.22% of 1125
(70.03 ÷ 3.03 × 12.02) ÷ 35.03 × 20.02 × 8.08 = ? × (9.09 2.02 – 1.01)Â
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exactvalue.)