[wolfgang]

A farmer has four kinds of animals with different numbers.

If we list them from higher number to lower number,they will be like this:

Pigs,,sheep,,cows,,horses

Number of pigs=P

Number of sheep=S

Number of cows=C

Number of horses=H

P<40

P+2C=3S=H^2

P+S+C=H^2+2

Find,P,S,C,and H

[curr3nt]

31, 27, 25 and 9

[k-man]

1) From P+2C=H² and P+S+C=H²+2 we get P+2C+2=P+S+C. Getting rid of P on both sides we get 2C+2=S+C and therefore S=C+2

2) Substituting S in P+2C=3S we get P+2C=3C+6 or P=C+6

3) Since 3S=H² and H<S<P<40, H must be divisible by 3, must be greater than 3 and less than sqrt(120). So H can only be 6 or 9.

If H=6 then S=12, C=10 and P=16

If H=9 then S=27, C=25 and P=31

Both answers satisfy the conditions of the puzzle.

[Guest]

And, since:

S=H^2/3

Thus, H can only be a multiple of 3.

Hence, plugging in the values of either H=6 or H=9 will in turn provide correct answers.

However, H cannot be 3, because it will make H = P = 3 and the constrain of the puzzle is:

P>S>C>H

Same happens with H=12, which will give an answer of P = 52, but then again, it's a matter of restrictions:

P<40

[Guest]

31, 27, 25 & 9

[wolfgang]

1) From P+2C=H² and P+S+C=H²+2 we get P+2C+2=P+S+C. Getting rid of P on both sides we get 2C+2=S+C and therefore S=C+2

2) Substituting S in P+2C=3S we get P+2C=3C+6 or P=C+6

3) Since 3S=H² and H<S<P<40, H must be divisible by 3, must be greater than 3 and less than sqrt(120). So H can only be 6 or 9.

If H=6 then S=12, C=10 and P=16

If H=9 then S=27, C=25 and P=31

Both answers satisfy the conditions of the puzzle.

Very good !