## Task $1 / 86$

We are looking for the smallest natural number $n$, which is the product of 3 prime factors $p_{1} ; p_{2} ; p_{3}$, and it holds that: $p_{3}=55 \cdot p_{1} \cdot p_{2}+1$ and $p_{3}>p_{2}>p_{1}$.
Certainly, here is the translation:

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Certainly, here is the translation:

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With certainty, $p_{3}>3$. Thus, $p_{3}$ is odd, and it follows that $p_{3}-1=55 \cdot p_{1} \cdot p_{2}$ is even. Therefore, at least one of the two primes $p_{1}$ and $p_{2}$ must be even; W.L.O.G. (due to the symmetry in $p_{1}$ and $p_{2}$), let this be $p_{1}=2$. This gives $p_{3}=110 \cdot p_{2}+1$.

The product $p_{1} p_{2} p_{3}$ is minimized when each of the individual factors is minimized. We therefore start by testing the primes $p_{2}$, beginning with $p_{2}=2$, to see if $110 \cdot p_{2}+1$ yields a prime number. For $p_{2}=2$, we get $p_{3}=221=13 \cdot 17$, which is not a prime. However, $p_{2}=3$ already yields the prime $p_{3}=331$. The desired product is thus $p_{1} p_{2} p_{3}=2 \cdot 3 \cdot 331=1986$.