Hint: the number $ac=10a+c $ is equal to $ac=10a+c=(9+1)a+c=9a +(a+c) $ So $ac =10a+c$ divided by 9, will have same remainder as $a+c$ divided by 9.
This is assuming you know the elementary school trick that a number is divisible by 9, if and only if the sum of its digits is divisible by 9.
And then it assumes you realize that that means the divisor when divided by 9 is going to have the same divisor as the sum of the digits divided by 9. Which we can figure out by adding those digits.
Ultimately the divisor is exactly equal to the sum of the sum of the sum of .... the digits.
So add up the numbers 1+2+3+4...... that's long but play with it and see if you can find shortcuts. Example: if 2+7=9 eventually in a future step you will want the remainder after dividing by 9 so that will be 0. So you can toss out any a+b=9. Keep playing and see what happens.
Now... you are simply taking the word of some stranger on the Internet the the remainder when divided by 9 is the same thing as the sum of the digits. Play with it and try to convince yourself it is true.
Hint: the number ac is 10a + c. 10 is (9+1) so ac=10a + c = 9a + (a+c). So ac and a+c will have the same divisor when divided by 9.