495. Find all pairs of natural numbers if their sum is 60, and the least common multiple is 72.
$\triangle$ Let the required numbers be $a$ and $b$.

The number 72 is divisible by each of them. Since

$$
72=8 \cdot 9=2^{3} \cdot 3^{2}
$$

the prime factorization of numbers $a$ and $b$ contains only twos and threes. Moreover, in the factorization of one of the numbers $a$ and $b$, the number 2 enters in the third power, and in the factorization of the other, only in the second power, because the sum $a+b=60$ is divisible by 4 but not by 8. For the same reason, the number 3 enters in the factorization of one of the numbers $a$ and $b$ in the second power, and the other only in the first.

Let's consider all cases.

1) Suppose

$$
a=2^{3} \cdot 3^{2}=72, \quad b=2^{2} \cdot 3=12
$$

This option does not work because the equality $a+b=60$ is not satisfied.

2) Suppose

$$
a=2^{2} \cdot 3=12, \quad b=2^{3} \cdot 3^{2}=72
$$

This case is essentially the same as the first.

3) Suppose

$$
a=2^{3} \cdot 3=24, \quad b=2^{2} \cdot 3^{2}=36
$$

Here, the condition $a+b=60$ is satisfied.

4) Suppose

$$
a=2^{2} \cdot 3^{2}=36, \quad b=2^{3} \cdot 3=24
$$

This option is the same as the third.

Answer: 24 and 36.