The average atomic weight of lead found in nature is 207.2. This (largely) consists of a mixture of three types of isotopes, which are the end products of different radioactive decay series, with atomic weights of 206, 207, and 208. In the mixture, there is approximately the same amount of 208 atoms as there are 206 and 207 atoms combined.

Express the approximate ratio of the number of the three types of lead atoms in the mixture using the smallest integers possible.
Let the number of atoms with a unit weight of 206 and 207 (measured in a suitable unit) in a certain quantity of lead be $x$ and $y$, respectively, then the number of 208s is $x+y$, and the average atomic weight (the weight per atom) of the respective quantity is

$$
\frac{206 x+207 y+208(x+y)}{2(x+y)}=207.2
$$

The numerator can be transformed as: $2 \cdot 207(x+y)+y$, thus, subtracting 207 from both sides

$$
\frac{y}{2(x+y)}=0.2, \quad 0.6 y=0.4 x
$$

which means $x: y=3: 2$. According to this, assuming the equality of the sum of the number of 206 and 207 isotopes and the number of 208s to be exact, the ratio of the numbers of the three isotopes is $3: 2: 5$.

Remark. According to the Encyclopedia of Natural Sciences (Akadémiai Kiadó, Budapest, 1968, Vol. 5, pp. 25-26), the three ratios in percentage are $23.6, 22.6, 52.3$; however, a small amount (1.5%) of 204 isotope is also present in natural lead. The second sentence in parentheses in the problem statement refers to this.