13.372 A consignment shop has accepted for sale cameras, watches, fountain pens, and receivers for a total of 240 rubles. The sum of the prices of a receiver and a watch is 4 rubles more than the sum of the prices of a camera and a fountain pen, and the sum of the prices of a watch and a fountain pen is 24 rubles less than the sum of the prices of a camera and a receiver. The price of a fountain pen is an integer number of rubles, not exceeding 6. The number of cameras accepted is equal to the price of one camera in rubles, divided by 10; the number of watches accepted is equal to the number of receivers, as well as the number of cameras. The number of fountain pens is three times the number of cameras. How many items of the specified types were accepted by the shop?
Solution. Let $x$ be the price of a watch, $y$ be the price of a pen, $z$ be the price of a receiver, $t$ be the price of a camera, and $n$ be the number of cameras. Then, according to the problem, we have:

$\left\{\begin{array}{l}t=10 n, \\ y \leq 6, \\ z+x=y+t+4, \\ 3 n y+n x+n z+n t=240, \\ n, y \text { - integers }\end{array} \Rightarrow\left\{\begin{array}{l}y \leq 6, \\ t=10 n, \\ 3 n y+n(y+2 t+4)=240, \\ n, y \text { - integers }\end{array} \Rightarrow\right.\right.$

$\Rightarrow\left\{\begin{array}{l}5 n^{2}+n(y+1)=60, \\ y \leq 6, \\ n, y \text { - integers }\end{array} \Rightarrow\left\{\begin{array}{l}5 n^{2}60, \\ n \text { - integers }\end{array} \Rightarrow\left\{\begin{array}{l}n \leq 3, \\ n \geq 3\end{array} \Rightarrow n=3, \Rightarrow 6 n=18\right.\right.\right.$.

Answer: 18.

Note. In the condition of problem 13.372, there is redundant information that the sum of the prices of a watch and a pen is 24 rubles less than the sum of the prices of a camera and a receiver. In the solution provided above, we did without this condition.