On the even side of a street, the sum of the house numbers from one corner to the other is 78, and there are at least 5 houses on this segment. What could be the house number of the fourth house from the corner?
Solution. Let the house number of the house standing at the corner be $x$. We know that $x+(x+2)+\ldots+(x+2 k)=78$, where $k \geq 4$. Based on the formula for the sum of an arithmetic sequence, this gives us

$$
\begin{aligned}
\frac{(2 x+2 k)(k+1)}{2} & =78 \\
(x+k)(k+1) & =78=2 \cdot 3 \cdot 13
\end{aligned}
$$

Since $k \geq 4$, then $k+1=6,13,26,39$ or 78. Since $x \geq 2$, in the last four cases, the value of $(x+k)(k+1)$ would be too large. In the first case, $k=5, x=13-5=8$, and thus the house number of the fourth house is $8+3 \cdot 2=14$.