## Task 32/75

We are looking for the smallest prime number $p$ such that

$$
\begin{gathered}
p+1 \equiv 0(\bmod 2) ; \quad p+1 \equiv 0(\bmod 3) ; \quad p+1 \equiv 0(\bmod 4) \\
p+1 \equiv 0(\bmod 5) ; \quad p+1 \equiv 0(\bmod 6)
\end{gathered}
$$
Due to $p+4 \equiv 0(\bmod 5)$, the last digit of the desired prime number $p$ must be 1 (if it were 6, then $p$ would not be a prime number).

Thus, $p+1 \equiv 0(\bmod 2)$ is automatically satisfied, so this condition does not need to be considered further. From the condition $p+5 \equiv 0(\bmod 6)$, it follows that $p+5 \equiv p-1 \equiv 0(\bmod 6)$, which means that among the set of at most two-digit prime numbers, only the primes 31 and 61 are possible. For both, $p+2 \equiv 0(\bmod 3)$ holds, but only $p=61$ also satisfies the condition $p+3 \equiv 0(\bmod 4)$. Therefore, $p=61$ is the smallest prime number with the required properties.

## Solution by Wolfgang Moldenhauer:

As a result of the last two conditions, there exist natural numbers $m$ and $n$ such that $p+5=5n$ and $p+5=6m$ hold. Subtracting the first equation from the second yields

$$
1=6m-5n
$$

Thus, $5n \equiv -1 \bmod 6 \equiv 5 \bmod 6$, and $n \equiv 1 \bmod 6$. For $n=1$, we get $p=1$, which is not a prime number. For $n=7$, we get $p=31$, which violates the third condition.

However, for $n=13$, we get $p=61$, which, as a check shows, is a solution.

## Solution by Bruno Hanisch:

From the congruences $p+4 \equiv 0 \bmod 5$ and $p+5 \equiv 0 \bmod 6$, the congruences

$$
6p+24 \equiv 0 \quad \bmod 30 \quad ; \quad 5p+25 \equiv 0 \quad \bmod 30
$$

follow. By subtracting, we obtain $p-1 \equiv 0 \bmod 30$. Therefore, the condition $p=30k+1(k \in \mathbb{N})$ must be satisfied. The number $p_1=31$ does not satisfy the congruence $p_1+3 \equiv 0 \bmod 4$, whereas $p_2=61$ satisfies all the required congruences. Consequently, $p_2=61$ is the smallest prime number that satisfies all the required congruences.

Other prime numbers with these properties are 181, 241, 421, 541, 601, 661, ...

## Solution by Karl-Bernd Dinter:

From the congruences, we get

$$
\begin{array}{llllll}
p+1 \equiv 0 & \bmod 2 & \rightarrow & p-1=2a & \rightarrow & a=\frac{1}{2}(p-1) \\
p+2 \equiv 0 & \bmod 3 & \rightarrow & p-1=3b & \rightarrow & b=\frac{1}{3}(p-1) \\
p+3 \equiv 0 & \bmod 4 & \rightarrow & p-1=4c & \rightarrow & c=\frac{1}{4}(p-1) \\
p+4 \equiv 0 & \bmod 5 & \rightarrow & p-1=5d & \rightarrow & d=\frac{1}{5}(p-1) \\
p+5 \equiv 0 & \bmod 6 & \rightarrow & p-1=6e & \rightarrow & e=\frac{1}{6}(p-1)
\end{array}
$$

where $a, b, c, d, e$ are integers.

It follows immediately that $p-1$ must be a common multiple of $2, 3, 4, 5$, and 6. The least common multiple of these numbers is 60.

Indeed, $p=60+1=61$ is a prime number and thus the smallest prime number that has the required properties.