Find the greatest number 23a68b, which is divisible by 3

Solution

For option 1, the sum of the digits = 2 + 3 + 8 + 6 + 8 + 9 = 36 (Divisible by 3 and 9) For option 2, the sum of the digits = 2 + 3 + 9 + 6 + 8 + 5 = 33 (Divisible by 3 but not divisible by 9) For option 3, the sum of the digits = 2 + 3 + 9 + 6 + 8 + 8 = 36 (Divisible by 3 and 9) For option 4, the sum of the digits = 2 + 3 + 7 + 6 + 8 + 7 = 33 (Divisible by 3 but not divisible by 9) So, 239685 and 237687 are divisible by 3 but not divisible by 9 The greatest number among them = 239685 ∴ The greatest number is 239685 which is divisible by 3 but NOT divisible by 9