Let the four-digit number be $\overline{abcd}$, where:

• $a$ is the हजार's digit, so $a\in\{1,2,\dots,9\}$
• $b,c,d\in\{0,1,2,\dots,9\}$
• Since each of the first three digits is divisible by the last digit, we must have $d\neq 0$.

So $d\in\{1,2,\dots,9\}$, and we need: $d\mid a,\quad d\mid b,\quad d\mid c.$

We count possible choices for each fixed last digit $d$.

1. Count choices for $a,b,c$ for a fixed $d$

For fixed $d$:

• $a$ must be a nonzero digit divisible by $d$.
• $b,c$ can be any digit divisible by $d$, including $0$.

Choices for $a$

Among $1$ to $9$, the number divisible by $d$ is $\left\lfloor \frac{9}{d} \right\rfloor.$

Choices for $b$

Among $0$ to $9$, the multiples of $d$ are $0,d,2d,\dots,\left\lfloor \frac{9}{d} \right\rfloor d,$ so the number of choices is $\left\lfloor \frac{9}{d} \right\rfloor + 1.$ Same for $c$.

Hence for fixed $d$, number of valid numbers is $\left\lfloor \frac{9}{d} \right\rfloor \left(\left\lfloor \frac{9}{d} \right\rfloor +1\right)^2.$

2. Sum over all possible $d=1$ to $9$

Let $m_d=\left\lfloor \frac{9}{d} \right\rfloor$. Then total count is $\sum_{d=1}^9 m_d(m_d+1)^2.$

Now compute one by one:

$d=1$

$m_1=9$ Count: $9\cdot 10^2=900$

$d=2$

$m_2=4$ Count: $4\cdot 5^2=100$

$d=3$

$m_3=3$ Count: $3\cdot 4^2=48$

$d=4$

$m_4=2$ Count: $2\cdot 3^2=18$

$d=5$

$m_5=1$ Count: $1\cdot 2^2=4$

$d=6$

$m_6=1$ Count: $1\cdot 2^2=4$

$d=7$

$m_7=1$ Count: $1\cdot 2^2=4$

$d=8$

$m_8=1$ Count: $1\cdot 2^2=4$

$d=9$

$m_9=1$ Count: $1\cdot 2^2=4$

3. Add all counts

$900+100+48+18+4+4+4+4+4=1086.$

Final Answer

The total number of such four-digit numbers is $\boxed{1086}.$