## Task 23/90

We are looking for the smallest natural number $n>1000$ with the following properties:

1. The digit sequence of $n$ is symmetric.
2. The prime factorization of $n$ contains exactly two prime factors $p_{1}$ and $p_{2}$ in 1st power.
3. When divided by $p_{1}$, $p_{2}$ leaves a remainder of 5.
## Solution by Kitaktus:

The number $1991=11 \cdot 181$ with $p_{1}=181$ and $p_{2}=11$ satisfies all conditions of the problem because $181=16 \cdot 11+5$. We now show that this is the smallest solution.

Assume there is a smaller such number $n>1000$. Then $n$ is a four-digit number and has the decimal representation $1 y y 1$ with $y \in\{0,1, \ldots, 8\}$. It holds:

- $1001=7 \cdot 11 \cdot 13$. Excluded, as it has three prime factors with multiplicity 1.
- $1111=11 \cdot 101$. Excluded, as $11 \not \equiv 5(\bmod 101)$ and $101 \equiv 2 \not \equiv 5(\bmod 11)$.
- $1221=3 \cdot 11 \cdot 37$. Excluded, as it has three prime factors with multiplicity 1.
- $1331=11 \cdot 11 \cdot 11$. Excluded, as it has no prime factors with multiplicity 1.
- $1441=11 \cdot 131$. Excluded, as $11 \not \equiv 5(\bmod 131)$ and $131 \equiv 10 \not \equiv 5(\bmod 11)$.
- $1551=3 \cdot 11 \cdot 47$. Excluded, as it has three prime factors with multiplicity 1.
- $1661=11 \cdot 151$. Excluded, as $11 \not \equiv 5(\bmod 151)$ and $151 \equiv 8 \not \equiv 5(\bmod 11)$.
- $1771=7 \cdot 11 \cdot 23$. Excluded, as it has three prime factors with multiplicity 1.
- $1881=3^{2} \cdot 11 \cdot 19$. Excluded, as $11 \not \equiv 5(\bmod 19)$ and $19 \equiv 8 \not \equiv 5(\bmod 11)$.

Therefore, there is no number $<1991$ that satisfies the conditions of the problem. 1991 is thus the smallest such number.

Note:

Condition 2 is ambiguously formulated. Does it mean that $n$ has exactly two prime divisors and these appear with multiplicity 1, or that $n$ has exactly two prime divisors with multiplicity 1, but possibly other prime divisors with higher multiplicity?

In the first case, $1881=32 \cdot 11 \cdot 19$ is excluded, in the second case it is not.

In the first interpretation, it can also be shown without case distinction that only the value 9 for $y$ is possible.