## Task 17/81

Let $a$ be the 144-digit natural number that (in decimal notation) is represented exclusively by the digit 1. Provide 8 different prime factors of $a$!
Obviously, the sum of the digits of $a$ is 144 and the alternating sum of the digits is 0. It follows (from known divisibility rules) that $a$ is divisible by 3 and by 11 without remainder. Thus, two prime factors of $a$ are found: $p_{1}=3 ; p_{2}=11$.

The remaining 6 prime factors can be found using Fermat's little theorem. This theorem states: $x^{p-1} \equiv 1(\bmod \mathrm{p})$, if $p$ is a prime and $\operatorname{gg} T(x ; p)=1$. Specifically for $x=10$ it follows

$$
\begin{array}{rlrl}
10^{6} & \equiv 1(\bmod 7) & 10^{144}=\left(10^{6}\right)^{24} \equiv 1(\bmod 7) \\
10^{12} & \equiv 1(\bmod 13) & 10^{144}=\left(10^{12}\right)^{12} \equiv 1(\bmod 13) \\
10^{16} & \equiv 1(\bmod 17) & 10^{144}=\left(10^{16}\right)^{9} \equiv 1(\bmod 17) \\
10^{18} \equiv 1(\bmod 19) & 10^{144}=\left(10^{18}\right)^{8} \equiv 1(\bmod 19) \\
10^{36} & \equiv 1(\bmod 37) & 10^{144}=\left(10^{36}\right)^{4} \equiv 1(\bmod 37) \\
10^{72} & \equiv 1(\bmod 73) & 10^{144}=\left(10^{72}\right)^{2} \equiv 1(\bmod 73)
\end{array}
$$

However, $10^{144}-1=9 \sum_{i=0}^{143} 10^{i}=9 a$ (this follows from the sum formula for the finite geometric series

$$
\sum_{i=0}^{n-1} x \cdot q^{i}=x \cdot \frac{q^{n}-1}{q-1}
$$

with $x=1, q=10, n=144)$. Since 9 on the one hand and the primes $p_{3}=7, p_{4}=13, p_{5}=17, p_{6}=19$, $p_{7}=37$ and $p_{8}=73$ on the other hand are coprime, i.e., since $g g T\left(9, p_{i}\right)=1$ for $i=3 ; 4 ; \ldots ; 8$ holds, it follows that these primes are prime factors of $a$.