If the three digits of a three digit number are a, B and C respectively, and (a + B + C) can be divided by 9, prove that the three digit number must be divided by 9 The answer is this: if these three numbers are a, B and C, then the value of three digits is 100A + 10B + C = 99A + 9b + (a + B + C), where 99A, 9b and (a + B + C) can be divisible by 9, so this three digit must be divisible by 9 But I don't understand. Please explain in detail, I just don't understand why a + B + C becomes 100A + 10B + C = 99A + 9b + (a + B + C)

If the three digits of a three digit number are a, B and C respectively, and (a + B + C) can be divided by 9, prove that the three digit number must be divided by 9 The answer is this: if these three numbers are a, B and C, then the value of three digits is 100A + 10B + C = 99A + 9b + (a + B + C), where 99A, 9b and (a + B + C) can be divisible by 9, so this three digit must be divisible by 9 But I don't understand. Please explain in detail, I just don't understand why a + B + C becomes 100A + 10B + C = 99A + 9b + (a + B + C)

The hundred, ten and individual digits of this number are a, B and C respectively, which means that this number is 100A + 10B + C
For example, a number is 123, which is 1 * 100 + 2 * 10 + 3, just equal to the number 123. With your explanation later, it's easy to understand