How many $4$-digit numbers of the form $1a2b$ are divisible by $3$? How many $4$-digit numbers of the form $\overline{1a2b}$ are divisible by $3?$
Hello I am new here so I don’t really know how this works. I know that for something to be divisible by 3, you add the digits and see if they are divisible by $3$. So that means $3+a+b=6, 9, 12, 15, 18,$ or $21.$ I’m just confused about how to calculate the number of cases.

Giving you a hint :-
You got $3 + a + b = 6,9,12,15,18$ or $21$, which implies that $a + b = 3,6,9,12,15$ or $18.$ Now do Case-Work and find all possible $a,b$ which can satisfy these . This may take a bit of work.
$($For e.g. when $a + b = 3$ we have $(a,b) = (0,3),(1,2)(2,1)(3,0))$
Note that you forgot  the case when $3 + a + b = 3$, in that case $(a,b)$ = $(0,0)$.
Edit :- Keep in mind that $a,b$ are $1$-digit numbers . Hence if $a + b = 12$ , $(a,b) = (1,11)$ is not a solution, but $(a,b) = (3,9)$ is a solution .
