Problem 5. In the confectionery Coconut, there are tables and chairs grouped into sets as follows: several sets of one table with four legs and four chairs with three legs, and several sets of one table with three legs and four chairs with four legs. The total number of legs is 264. How many children can sit in the confectionery?
Solution. Let the number of sets of the first type be denoted by $a$, and of the second type by $b$. The total number of socks of the first type is $4+4 \cdot 3=16$, and of the second type is $3+4 \cdot 4=19$. Now, for the total number of socks, we form the equation $16a + 19b = 264$. From this, $19b = 264 - 16a = 8(33 - 2a)$. Since 8 divides the right side of the last equation, it is clear that it must also divide the left side. But 19 is not divisible by 8, so 8 must be a divisor of $b$, i.e., $b \in \{8, 16, 24, \ldots\}$. Meanwhile, $19b \leq 264$. Clearly, then $b=8$, and $a=7$. It follows that the number of free places in the bakery is 15$\cdot$4=60.

## VII Department