THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

For any choice of 6 distinct non-zero digits, at least two of them are consecutive. Indeed, if $a<b<c<d<e<f$ and no digits are consecutive, then $b\ge a+2$, $c\ge b+2\ge a+4$, $d\ge c+2\ge a+6$, $e\ge d+2\ge a+8$ and $f\ge e+2\ge a+10$, which is impossible, since both $a$ and $f$ are digits. Denote by $m$ and $n$, $m>n$, two consecutive digits among $a,b,c,d,e,f$, and by $p,q,r,s$ the other four digits. We have $x\mid \overline{pqrsmn}$ and $x\mid \overline{pqrsnm}$, hence $x\mid \overline{pqrsmn}-\overline{pqrsnm}=9(m-n)$. Since $m-n=1$, we find $x\mid 9$. If $3\nmid a+b+c+d+e+f$, the only possibility is $x=1$. If $3\mid a+b+c+d+e+f$ and $9\nmid a+b+c+d+e+f$, then $x=1$ or $x=3$. If $9\mid a+b+c+d+e+f$, then $x=1,x=3$ or $x=9$.