Different Stools For Different People

Let $a$ and $b$ denote the number of people sitting on a 3-legged stool and 4-legged stool respectively.

Then once each people sit on each stool, a 3-legged stool have 5 legs touching the floor; and a 4-legged stool have 6 legs touching the floor.

Solve the Diophantine equation $5a+6b=39$ and get $a=-39+6n$ and $b=39-5n$ . Then, $a+b=n$ , and the only value of $n$ which makes both $a$ and $b$ positive is $n=7$ , as it is between $39/5$ and $39/6$

The three-legged stools can be counted as five -legged due to the people on them, and the four-legged ones can be counted as six -legged. We can then count further and say that we only have five-legged chairs and stools, with a few "extra" legs left over. We search for the multiple of five closest to $39$ and choose $35$ . That's a total of $4$ left-over legs, so:

Method 1:

• There are $4$ six-legged $=$ four-legged chairs

• And $\frac{39-24}{5}=3$ five-legged $=$ three-legged stools.

Method 2:

We have $\frac{35}{5}=7$ chairs and stools overall.

The sum legs of each stool is 5(3+2) and of each chair is 6(4+2).Total sum is 39,so we try the sum of different numbers and find : 3x5+4x6=39,so a=3 and b=4,a+b=7

So somehow you have to know that all those people have their two legs on the floor? I rarely sit like that. My legs can't even touch the floor on some stools.

$5+6 = 11 \\ 39 \equiv 6 (mod\,11)$

So, 11 is three times in 39 and leave a rest of six which is equivalent to another chair 4-legged.

Therefore

$3\times 2 = 6; \\ 6+1=7$