154. A five-digit number is divisible by 72, and three of its digits are ones. Find all such numbers.
Let's denote two unknown digits of a five-digit number by $x$ and $y$. One of them, for example, $y$, stands at the last place, since 1 cannot be the last digit of an even number.

We apply the divisibility rule for 9:

$$
(1+1+1+x+y): 9, \quad(3+x+y): 9
$$

From this, $x+y=6$ or $x+y=15$.

There is a problem with the position of the other unknown digit $x$: it can occupy the first, second, third, or fourth place. Let's consider all four cases.

1) Suppose the digit $x$ stands in the number at the first place. Then the number takes the form $\overline{x 111 y}$.

By the divisibility rule for 8, the number $\overline{11 y}$ is divisible by 8, hence $y=2$.

Since $x+y=6$ or $x+y=15$, then respectively $x=4$ or $x=13$. But the second possibility is invalid.

2) Suppose the digit $x$ stands at the second place.

The reasoning from the first case applies here as well, so $y=2, x=4$.

3) Suppose the digit $x$ occupies the third place. Then the number has the form $\overline{11 x 1 y}$.

By the divisibility rule for 8, the number $\overline{x 1 y}$ is divisible by 8. Let's transform this three-digit number:

$$
\begin{aligned}
\overline{x 1 y}=100 x+10+y= & (8 \cdot 12 x+4 x)+(8+2)+y=(8 \cdot 12 x+8)+(4 x+y+2)= \\
& =(8 \cdot 12 x+8)+(x+y)+(3 x+2)
\end{aligned}
$$

(here the term equal to $x+y$ is highlighted).

If $x+y=6$, then

$$
(6+(3 x+2)): 8, \quad(8+3 x): 8, \quad 3 x: 8, \quad x: 8
$$

Therefore, $x=0$ or $x=8$. If $x=0$, then $y=6$, and if $x=8$, then $y=-2$, which is impossible.

If $x+y=15$, then

$$
(15+(3 x+2)): 8, \quad(17+3 x): 8, \quad(16+(3 x+1)): 8, \quad(3 x+1): 8
$$

Then $x=5$, from which $y=10$, which is also impossible.

4) Suppose the digit $x$ in the number occupies the fourth place. The number takes the form $\overline{111 x y}$.

The number $\overline{1 x y}$ must be divisible by 8. Let's perform similar transformations with it as in the previous case:

$$
\begin{aligned}
\overline{1 x y}=100+10 x+y= & (8 \cdot 12+4)+(8 x+2 x)+y=(8 \cdot 12+8 x)+(2 x+y+4)= \\
& =(8 \cdot 12+8 x)+(x+y)+(x+4)
\end{aligned}
$$

If $x+y=6$, then

$$
(6+x+4): 8, \quad(x+10): 8, \quad((x+2)+8): 8, \quad(x+2): 8
$$

From this, $x=6, y=0$.

If $x+y=15$, then

$$
(15+x+4): 8, \quad(x+19): 8, \quad(x+3): 8
$$

Then $x=5, y=10$, which is impossible.

Answer: $41112,14112,11016,11160$.