385. Three people want to buy a house for 26000 pounds. They agreed that the first would give half, the second one third, and the third one quarter. How much will each give?

## Leibniz's Problems.
385. Since $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$ and this amounts to 26000, then $\frac{1}{12}$ equals 2000, therefore, the first gives 12000, the second 8000, and the third 6000 livres.

Leibniz, Gottfried Wilhelm (1646-1716), a famous lawyer, philosopher, and mathematician. Leibniz, along with Newton, is often credited as the founder of infinitesimal analysis. This is true in the sense that these brilliant scientists systematized the ideas and results obtained by their predecessors and, by introducing uniformity in the methods and notations of analysis, created a coherent, practically convenient, and powerful apparatus. They also enriched analysis with a number of new and important achievements. Many of Leibniz's symbolic notations have persisted in science to this day, for example, the signs of the integral and differential belong to him. He worked on the study of the properties of curves (for example, the catenary), on the expansion of functions into series, and finally, from him, Serret began the idea of determinants.

386 1) $m^{5}-m=m\left(m^{4}-1\right)=m\left(m^{2}-1\right)\left(m^{2}+1\right)$.

Let $m=2 k$, i.e., even. Then $m^{2} \pm 1=4 k^{2} \pm 1$; if $m$ is not divisible by 5, then the possible forms of this number are $5 n \pm 1$ and $5 n \pm 2 ; m^{2}$ gives $25 n^{2} \pm 10 n+1$ and $25 n^{2} \pm 20 n+4$, i.e., $m^{2}=$ mod $5+1$ or mod $5-1$, therefore, either $m^{2}+1$ or $m^{2}-1$ is divisible by 5. Similarly, we can verify this when $m$ is odd.
2) $m^{7}-m=m\left(m^{6}-1\right)=m\left(m^{3}-1\right)\left(m^{3}+1\right)$.

Let $m=2 k$, i.e., even. For $m$, if it is not divisible by 7, the possible forms are $7 n \pm 1, 7 n \pm 2, 7 n \pm 3$. But $m^{3}$ in all these cases will be equal to mod $7 \pm 1$, therefore, either $m^{3}-1$ or $m^{3}+1$ is divisible by 7.