7 horses

72 chickens

21 goats

70 +9+21 = 100

]]>h = no. of horses, g = no. of goats, d = no. of bunches of ducks.

So

h + g + 8d = 100 (count restriction) => h + 8d = 100 – g

10h + g + d = 100 (cost restriction) => 10h + d = 100 – g

Therefore

h + 8d = 10h + d => 9h = 7d

We cant have a fraction of a horse or a number of bunches of ducks that is not a multiple of 8

i.e. 3 horses would give us 3×9/7 = 3.87 bunches of ducks

or 6 bunches of ducks would give us 6×7/9 = 4.66 horses.

So we need a multiple of the lowest common denominator of 9 and 7 that is less than 100 which is 56.

So we can only have 7 horses and 9 bunches of ducks.

From a count point of view, that is 7 + (9 x 8) = 79

From a cost point of view, that is (7 x 10) + 9 = 79

and since a goat costs 1 and counts 1, 100 – 79 = 21

7 Horses, 21 Goats and 72 ducks.

]]>72 Ducks,

21 Goats.

Thank god for that I was getting stressed ]]>

Answer 1: gravity, throw it up, because watch goes up must come down.

Answer 2: 9 horses, 9 goats and 8 ducks.

But alas, that seems too easy, so I’m probably wrong. ]]>

Second one: 7 horse for €70. 21 goats for €21 and 72 ducks for €9.

7 horse + 21 goats + 72 ducks = 100 animals

€70 + €21 + €9 = €100

:)

]]>80 ducks, 10 goats, 1 horse.

]]>9 horses, 9 goats, and 8 ducks is one solution. ]]>