When a positive integer x is divided by 9, the remainder is 6, and when divided by 21, the remainder is 12. If the number lies between 300 and 350, then the sum of the digits of x is:

(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