24. Six integers. Find six different smallest integers in such a way that the product of any five arbitrarily chosen from this six equals one or several periods in the decimal representation of the number, which is the reciprocal of the remaining sixth number. For example, if we take the decimal representation of the reciprocal of 41,

$$
\frac{1}{41}=0,243902439 \ldots
$$

then the period will be 02439.
24. If $b$ is the period in the decimal representation of the number reciprocal to $a$, then all digits of the product $a b$ are nines. Further, $99=$ $9 \cdot 11, 999=3 \cdot 9 \cdot 37, 9999=9 \cdot 11 \cdot 101, 99999=0 \cdot 41 \cdot 271$ and $999999=3 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 37$. The last six numbers are the smallest distinct integers that satisfy the given condition. Indeed,

$$
\begin{array}{cc}
\frac{1}{3}=0.333333 \ldots & \frac{1}{11}=0.090909 \ldots \\
\frac{1}{7}=0.142857 \ldots & \frac{1}{13}=0.076923 \ldots \\
\frac{1}{9}=0.111111 \ldots & \frac{1}{37}=0.027027 \ldots \\
& {[\text { M. M., 35, 311 (November 1962).] }}
\end{array}
$$