## Task $7 / 78$

The smallest natural number of the form $2^{n}-1$ (with $n \in N, n>0$) that is divisible by 1001 without a remainder is sought.
Due to $1001=7 \cdot 11 \cdot 13$, $2^{n}-1$ must be divisible by 7, 11, and 13 without a remainder. In other words, it must hold that

$$
2^{n} \equiv 1(\bmod 7) ; \quad 2^{n} \equiv 1(\bmod 11) ; \quad 2^{n} \equiv 1(\bmod 13)
$$

Now, $2^{3}=8 \equiv 1(\bmod 7)$; thus, $n \equiv 0(\bmod 3)$. Furthermore, $2^{5}=32 \equiv-1(\bmod 11)$, hence $2^{10} \equiv 1(\bmod 11)$ and therefore $n \equiv 0(\bmod 10)$.

Finally, $2^{6}=64 \equiv-1(\bmod 13)$, so $2^{12} \equiv 1(\bmod 13)$, thus $n \equiv 0(\bmod 12)$. From all this, it follows that $n \equiv 0(\bmod 60)$ and due to the minimality requirement, $n=60$.

The number $2^{50}-1=1152921504606846975$ is therefore the smallest natural number of the form $2^{n}-1$ that is divisible by 1001 without a remainder.