## Task 36/80

The smallest natural number $n$ is sought for which the following holds

$$
n \equiv 1(\bmod 2) ; \quad n \equiv 1(\bmod 3) ; \quad n \equiv 1(\bmod 5) ; \quad n \equiv 0(\bmod 7) ; \quad n \equiv 1(\bmod 11)
$$
Due to $n \equiv 0(\bmod 7)$, $n$ can be represented in the form $n=7k, k \in N$. Because

$$
\begin{array}{r}
n \equiv 1(\bmod 2) \text { and } 7 \equiv 1(\bmod 2) \text {, } k=1(\bmod 2) \\
n \equiv 1(\bmod 3) \text { and } 7 \equiv 1(\bmod 3) \text {, } k=1(\bmod 3) \\
n \equiv 1(\bmod 5) \text { and } 7 \equiv 2(\bmod 5) \text {, } k=3(\bmod 5) \\
n \equiv 1(\bmod 11) \text { and } 7 \equiv 7(\bmod 11) \text {, } k=8(\bmod 11)
\end{array}
$$

We assume that the smallest number $k$ is at most three digits:

$$
k=100a_{2}+10a_{1}+a_{0}
$$

with $0 \leq a_{0} ; a_{1} ; a_{2} \leq 9, a_{1} ; a_{2} ; a_{3} \in N$. Then, due to (3), $a_{0}=3$ or $a_{0}=8$, and due to (1), $a_{1} \neq 8$, so it follows that $a_{1}=3$. According to (4), we then have

$$
a_{2}-a_{1}+3=11s+8 \quad ; \quad a_{2}-a_{1}=11s+5
$$

$(s \in G)$. From $0 \leq a_{1} ; a_{2} \leq 9$ it follows that $-9 \leq a_{2}-a_{1}=11s+5 \leq 9$. These inequalities are only satisfied for $s=0$ and for $s=-1$. Thus, from (5) we get $a_{2}=a_{1}+5 \geq 5$ or $a_{2}=a_{1}-6 \leq 3$.

Since we are looking for the smallest number $k$, we first try the smaller values for $a_{2}$, i.e., $a_{2}=0 ; 1 ; 2 ; 3$. This results in $a_{1}=6 ; 7 ; 8 ; 9$ and $k=63 ; 173 ; 283 ; 393$. Of these $k$-values, only $k=283$ satisfies condition (2). From this, we get $n=7 \cdot 283=1981$.

### 2.21 Problems and Solutions 1981