430. A man, on his death, left a will according to which the eldest son receives 100 francs and one-tenth of the remainder, the second son - 200 francs and one-tenth of the new remainder, the third - 300 francs and one-tenth of the new remainder, and so on until the last. In this way, the shares of all the sons should be equal. Find the size of the inheritance left, the number of sons, and the share of each.

## Francker's Problems.
430. Let the entire inheritance be $x$ francs. The first one, according to the condition, receives

$100+\frac{1}{10}(x-100)=\frac{1}{10} x+9$.

The remainder will be $\frac{9}{10} x-90$. The second one receives

$$
200+\frac{1}{10}\left(\frac{9}{10} x-290\right)
$$

But the shares of the first and second are equal, therefore,

$$
\frac{1}{10} x+90=171+\frac{9}{100} x
$$

from which $x=8100$. Each share in this case is 900 francs, and the number of sons is 9.

Franeker, a French mathematician, professor at the "Faculté des sciences" in Paris, author of "A Complete Course of Pure Mathematics," which was widely distributed and translated into Russian.