It is proved that the sum of the numbers on each digit of a number divisible by 3 can be divisible by 3

It is proved that the sum of the numbers on each digit of a number divisible by 3 can be divisible by 3

Suppose there is a four digit ABCD, which can be expressed in the following form: ABCD = 1000A + 100b + 10C + D = 999a + 99b + 9C + A + B + C + D = 9 × (111a + 11b + C) + A + B + C + D. We can see that 9 × (111a + 11b + C) must be divisible by 3, so to judge whether ABCD can be divisible by 3, we can see that a + B + C + D can be divisible by 3, that is to say