In a running race, 12 runners, marked with numbers, started. What was the order of arrival if the product of the bib number and the finishing position was always 1 greater than a number divisible by 13.



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I. Solution: Let's denote the number on the card by $r$ and the position number by $h$. We can write the Diophantine equation

$$
r h=13 u+1
$$

where, according to the problem, $0<r<13$, and $0<h<13$.

Let's examine $h$ for different values of $r$.

$$
\begin{aligned}
& r=1, \quad h=13 u+1, \quad \text { the appropriate } u=0, \quad \text { and thus } h=1, \\
& r=2, \quad h=\frac{13 u+1}{2}, \quad " \quad \text { " } \quad u=1, \quad \text { " } \quad h=7 \\
& r=3, \quad h=\frac{13 u+1}{3}, \quad " \quad \prime \quad u=2, \quad \text { " } \quad h=9 \\
& r=4, \quad h=\frac{13 u+1}{4}, \quad " \quad \text { " } \quad u=3, \quad \text { " } \quad h=10 \\
& r=5, \quad h=\frac{13 u+1}{5}, \quad " \quad \text { " } \quad u=3, \quad \text { " } \quad h=8 \\
& r=6, \quad h=\frac{13 u+1}{6}, \quad " \quad \prime \quad \quad u=5, \quad \prime \prime \prime h=11
\end{aligned}
$$

Further calculations are unnecessary because if $r$ and $h$ are suitable numbers, then $13-r$ and $13-h$ are also suitable, because $(13 r)(13-h)=13^{2}-13(r+h)+r h$, from which it is clear that $(13-r)(13-h)$ divided by 13 gives the same remainder as $r h$.

According to these, the order of arrival according to the numbers on the cards is:

$1,7,9,10,8,11,2,5,3,4,6,12$.

Tamás Gyula (Bp. II, Rákóczi g. I. o. t.)

II. Solution: The product of the number on the card and the position number can be at most $12 \cdot 12=144$. Let's write down the numbers of the form $13 k+1$ up to 144: 1, 14, 27, 40, 53, 66, 79, 91, 105, 118, 131, 144. The numbers 53, 79, 91, 105, 118, 131 cannot be considered because they are either prime numbers or one of their factors is greater than 12. The remaining numbers can be factored into two factors, each of which is at most 12:

$$
\begin{aligned}
1 & =1 \cdot 1 \\
14 & =2 \cdot 7 \\
27 & =3 \cdot 9 \\
40 & =\left\{\begin{array}{l}
4 \cdot 10 \\
5 \cdot 8
\end{array}\right. \\
66 & =6 \cdot 11 \\
144 & =12 \cdot 12
\end{aligned}
$$

From this, it is clear that the order of arrival according to the numbers on the cards was:

$1,7,9,10,8,11,2,5,3,4,6,12$