Problem 7.5. A store sells four types of nuts: hazelnuts, almonds, cashews, and pistachios. Stepan wants to buy 1 kilogram of nuts of one type and another 1 kilogram of nuts of a different type. He calculated how much such a purchase would cost him depending on which two types of nuts he chooses. Five out of six possible purchases by Stepan would cost 1900, 2070, 2110, 2330, and 2500 rubles. How many rubles does the cost of the sixth possible purchase amount to?
Answer: 2290.

Solution. Let $a, b, c, d$ be the cost of 1 kilogram of hazelnuts, almonds, cashews, and pistachios, respectively. From the condition, it follows that the set $A=\{1900,2070,2110,2330,2500\}$ is contained in the set $B=\{a+b, b+c, c+d, d+a, a+c, b+d\}$.

Note that the 6 elements of set $B$ can be divided into 3 pairs $(a+b, c+d),(b+c, d+a)$, $(a+c, b+d)$ with the same sum $a+b+c+d$. This means that in set $A$, 2 pairs of numbers with the same sum can be identified.

It is not difficult to understand that these can only be the pairs $(1900,2500)$ and $(2070,2330)$ with a sum of 4400 (for example, one can notice that all other pairs either intersect or their sums differ in the last two digits). Then the unknown sixth cost is easily calculated: $4400-2110=2290$ rubles.

Remark. The condition of the problem is satisfied for $a=930, b=970, c=1140, d=1360$.