Count the 4-digit integers that are multiple of 9 I want to count the total number of 4-digit integers that are multiple of 9, without any zero digit.
I was wondering if the best strategy is to treat it as a count problem or just apply some properties from number theory.
Naive way is to solve $a+b+c+d = 9, 18, 27, 36$ with $1 \leq a, b, c, d \leq 9$ but I feel this takes much time with too many case works to consider.
Any way we could do faster?

Okay, so you have determined that $abcd = 1000a + 100b + 10c + d$ is a multiple of $9$ if and only if $a+b+c+d$ is a multiple of $9$.
But now the trick is to realize that $a+b+c+d$ is a multiple of $9$ if and only if
$a+b+c+d \equiv 0 \pmod 9$ if and only if
$d\equiv -(a+b+c) \pmod 9$.
Now for any possible value $a+b+c$ may take, $-(a+b+c)\equiv$ to a specific digit $\pmod 9$ and there is exactly 1 digit between $1$ and $9$ that $d$ can take to be equivalent to $-(a+b+c)$.
(For example, if I just randomly pick $a= 5; b=7; c=2$ then $a+b+c\equiv 5\pmod 9$ and in order to have $572d$ divisible by $9$ we must have $a+b+c + d\equiv 5+d\equiv 0 \pmod 9$ or in other words $d \equiv -5 \equiv 4\pmod 9$.  We must have $d=4$.  It is the only choice and whatever choices we have for $a,b,c$ there will always bu exactly one choice for $d$).
So the number of four digit numbers equals
$(\text{number of choices for }a)\cdot(\text{number of choices for }b)\cdot(\text{number of choices for }c)\cdot(\text{number of choices for }d)=$
$9\cdot 9\cdot 9\cdot 1 = 9^3=729$
