When a positive integer x is divided by 9, the remainder

Solution

(9x + 6) __________ (1) Then, we put the values in place of x as (33,34,35,36,37,38) is the value of x cab be come in between 300 to 350. When we put x = 33, in eq(1) We get = 9 x (33) + 6 = 303 When we put x = 34, in eq(1) We get = 9 x (34) + 6 = 312 When we put x = 35, in eq(1) We get = 9 x (35) + 6 = 321 When we put x = 36, in eq(1) We get = 9 x (36) + 6 = 330 When we put x = 37, in eq(1) We get = 9 x (37) + 6 = 339 When we put x = 38, in eq(1) We get = 9 x (38) + 6 = 348 Hence, let another equation can be formed as (21x + 12) ________ (2) Then, we put the values in place of x as (14,15,16) When we put x = 14, in eq(2) We get = 21 x (14) + 12 = 306 When we put x = 15, in eq(2) We get = 21 x (15) + 12 = 327 When we put x = 16, in eq(2) We get = 21 x (16) + 12 = 348 Now x = 348 is the number  which lies between 300 and 350 So, the sum of digit (348) is = 3 + 4 + 8 = 15