Why is the sum of numbers divisible by 9 a multiple of 9

Why is the sum of numbers divisible by 9 a multiple of 9

Let y = A0 + A1 * 10 + A2 * 10 ^ 2 + +An * 10 ^ n because the remainder of 10 ^ X / 9 is always 1, that is, the remainder of AI * 10 ^ n / 9 is always AI (I ∈ n), so the remainder of each term of Y divided by 9 is always AI, so y can be divisible by 9, that is, the sum of the remainder of each term of y can be divisible by 9, y '= A0 + A1 + A2 + +...

The () of each digit of a number can be divided by (), which is a multiple of 3

The sum of each digit of a number can be divided by (3), which is a multiple of 3

A four digit number can be divided by 9, and the three digit number obtained after removing the last digit is just a multiple of 4. The last digit of the largest of the four digits is ()
A. 6B. 4C. 3D. 2

∵ the greatest multiple of 4 in three digits is 996, and ∵ the condition that four digits can be divided by 9 is that the sum of four digits is a multiple of 9, ∵ 9 + 9 + 6 = 24, ∵ if it is a multiple of 9, the last digit can only be 3

If a four digit number can be divided by 9, and the three digit number obtained after removing the last digit is just a multiple of 4, then the last digit of the largest of the four digits is______ ．

The maximum number of four digits should be 9. Therefore, the maximum number of four digits satisfying the condition can be set as 99ab. Because 99A can be divided by 4, the maximum value of a is 6. Because 99ab can be divided by 9, B = 3, so the answer is 3