Then the result is (10x+y)-(x+y)=9x.
The result must be a multiple of 9. Obviously, only option d meets the problem.
This is a senior high school group. I am a junior high school student, and I know something about it. If the first two are AB, then there are two.
If the front two are AC, then there are two, if the front is BA, then there are two, if the front is BB, then there are two, if the front is BC, then there are three, if the front is CA, then there are two, if the front is CB, then there are three, if the front is CC, then there are three, a total of 19. Mine may not be as simple as that just now.
Observe in groups of four.
n？ —(n+ 1)？ —(n+2)？ +(n+3)？ =-(2n+ 1)+2n+5=4
Therefore, every four consecutive natural numbers, the sum of this form is a constant value, that is, 4.
There are 6 groups of 4 consecutive natural numbers.
So, 1? —2? —3? +4? +5? —6? —7? +8? +……+24?
=4×(24÷4)
=24