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'.
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.
18.002 × 109.995 + 1499.996% of 450.08 + 12.005 % of 2999.997 = ? × 30.008 × 25.009
5275 of 105% + 99.07 × 17.889 =?
...? = 54.89 × 270.08 ÷ 135.17 + 464.35 ÷ 29.03
?% of (128.31 ÷ 15.97 × 75.011) = 419.98
13³ + 1.3² + 1.03¹ + 1.003 =?
2380.03 ÷ 84.98 x 39.9 = ? + 15.32
[(17.97)2 ÷ 47.67 X 11.67] ÷ 26.85 = ?2 - (10.98 X 65.98)
20.22% of (61.9 × 5.01) + 69.97 =?
(11.75)2 - 49.99% of 120 - ? = (8.23)2
149.78% of 319.87 – 199.83% of 45.45 = 130.03% of (? × 12.01)