Problem 5. In a room, there are chairs with 3 and 4 legs. When one person sits on each chair, there are a total of 69 feet in the room. How many chairs with 3 legs are there in the room, and how many with 4 legs?
Solution. Let the number of chairs with 3 legs be denoted by $a$, and the number of chairs with 4 legs by $b$. Then, according to the condition of the problem, we have: $3a + 4b + 2(a + b) = 69$. After simplification, we get $5a + 6b = 69$, i.e., $5a = 69 - 6b$. The right side of this equation must yield a natural number that is divisible by 5, i.e., a number that ends in the digit 0 or 5.

The number $69 - 6b$ cannot end in 0, because the number $6b$ can only end in one of the numbers $6, 2, 8, 4$, or 0.

Therefore, the number $69 - 6b$ must end in the digit 5, i.e., $6b$ must end in 4. The latter is possible if $b = 4, b = 9, b = 14, \ldots$. Clearly, $b < 12$, because $69 - 6b$ is a natural number.

Finally, $b = 4$ or $b = 9$, so $a = 9$ or $a = 3$, respectively.

## VIII Department