5. The giants were prepared 813 burgers, among which are cheeseburgers, hamburgers, fishburgers, and chickenburgers. If three of them start eating cheeseburgers, then in that time two giants will eat all the hamburgers. If five take on eating hamburgers, then in that time six giants will eat all the fishburgers. If seven start eating fishburgers, then in that time one giant can eat all the chickenburgers. How many burgers of each type were prepared for the giants? (The time it takes for one giant to eat one burger does not depend on the type of burger, and all giants eat at the same speed.)
Answer: 252 fishburgers, 36 chickenburgers, 210 hamburgers, and 315 cheeseburgers.

Solution: Let $a, b, c$, and $d$ be the quantities of cheeseburgers, hamburgers, fishburgers, and chickenburgers, respectively. According to the problem, $a + b + c + d = 813$. The statement that while three people eat cheeseburgers, two will eat all the hamburgers means that $\frac{a}{3} = \frac{b}{2}$. Similarly, the equalities $\frac{b}{5} = \frac{c}{6}$ and $\frac{c}{7} = d$ hold. Express all unknowns in terms of one, for example, $c$. We have $d = \frac{c}{7}, b = \frac{5c}{6}$, and $a = \frac{3b}{2} = \frac{5c}{4}$. Then $\frac{5c}{4} + \frac{5c}{6} + c + \frac{c}{7} = 813$, from which $c = 252$. Then $d = 36, b = 210$, and $a = 315$.

Comment: The answer is found by trial and error, and it is not shown that there are no other solutions - 3 points. $3$ arithmetic errors with correct reasoning reduce the score by 1-3 points.