752. Find the smallest natural number that is divisible by 28, and when divided by 15, gives a remainder of 4.
$\triangle$ The desired number is, on the one hand, $28 x$, and on the other $-15 y+4$, where $x$ and $y$ are natural numbers. We obtain the equation

$$
28 x=15 y+4
$$

We do not need to find all solutions to this equation in natural numbers, but only one solution - the one for which the values of $x$ and $y$ are minimal. For this purpose, we transform the equation using considerations of divisibility by 15:

$$
30 x-2 x=15 y+4, \quad 30 x-15 y=4+2 x
$$

from which

$$
(2 x+4): 15, \quad(x+2): 15
$$

Since we need the smallest value of $x$ that satisfies the last divisibility, we set $x+2$ equal to 15:

$$
x+2=15, \quad x=13 \text {. }
$$

Answer: 364.