Consider the following three arithmetic progressions:

$2,11,20,29, \ldots$

$7,15,23,31, \ldots$

$8,19,30,41, \ldots$

Determine the smallest (positive) common term of these three progressions!
Let $N$ be the common term of the three arithmetic sequences, such that

$$
N=2+9(x-1)=7+8(y-1)=8+11(z-1)
$$

Here $x, y, z$ represent the position of $N$ in each of the sequences.

$$
\begin{aligned}
& 2+9(x-1)=7+8(y-1) \\
& 2+9(x-1)=8+11(z-1) \\
& 8 y=9 x-6=8(x-1)+x+2 \ldots \\
& 11 z=9 x-4=11 x-2 x+2 \ldots
\end{aligned}
$$

From (1) it follows that $x+2$ is a multiple of 8;

From (2) it follows that $x+2$ is a multiple of 11.

Therefore, $x+2$ must be a multiple of 88:

$$
x+2=88 n, \quad x=88 n-2
$$

and thus

$$
y=99 n-3, \quad z=72 n-2
$$

The smallest common term value is obtained when $n=1$, so

$$
x=86, \quad y=96, \quad z=70 \quad \text { and then } \quad N=767
$$

The smallest common term is the 86th in the first sequence, the 96th in the second sequence, and the 70th in the third sequence.

Kornis Edit (Ráskai Lea g. V. o. Bp. V.)