Someone says that seven times their birth year, when divided by 13, leaves a remainder of 11, and thirteen times their birth year, when divided by 11, leaves a remainder of 7. In which year of their life will this person be in 1954?
Let the birth year be denoted by $x$, then according to the problem,

$$
\begin{aligned}
7 x & =13 y+11 \\
13 x & =11 z+7
\end{aligned}
$$

Subtracting 7 times (2) from 13 times (1), we get

$$
169 y-77 z+94=0
$$

Thus,

$$
\begin{aligned}
& z=\frac{169 y+94}{77}=2 y+1+\frac{15 y+17}{77}=2 y+1+u \\
& y=\frac{77 u-17}{15}=5 u-1+\frac{2 u-2}{15}=5 u-1+2 v \\
& u=15 v+1
\end{aligned}
$$

By substituting back,

$$
y=75 v+5-1+2 v=77 v+4
$$

and thus

$$
x=\frac{13 y+11}{7}=\frac{13 \cdot 77 v+52+11}{7}=143 v+9
$$

Given the nature of the problem,

$$
1750<x<1954
$$

therefore,

$$
1741<143 v<1945
$$

or

$$
12<v \leqq 13
$$

and thus

$$
v=13
$$

Therefore,

$$
x=143 \cdot 13+9=1868
$$

which means the person in question is currently in their 86th year of life.

Ferenc Nagy (Kecskemét, Szakérettsis School)