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I was sent this puzzle, I solved it by deduction and trial and error. It is keeping me awake at night trying to solve it with maths, any help appreciated.

Farmer Tony Smith bought 100 animals for his new farm.

He paid $3-00 each for the chickens, $8-00 each for the guinea pigs and $21-00 each for the geese.

Tony spent a total of $600-00
Tony could not remember how many of each animal he had bought.

However, he did remember that there was an even number of geese!

Can you work out how many of each animal Tony bought?

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Let’s designate the number of chickens, guinea pigs and geese by x, y and z, respectively.

x + y + z = 100 (1)
3x + 8y + 21z = 600 (2)

Subtract equation (1) three times from equation (2)

5y + 18z = 300

Since 5y and 300 are divisible by 5, z must be divisible by 5. And z is even, so z must be divisible by 10.

Solution 1: z = 0, so 5y = 300 from which follows y = 60 and since the total is 100, x = 40.
40 chickens = $120, 60 guinea pigs = $480, 0 geese = $0, total $600.
If you assume that he bought at least one of each, this solution is rejected.

Solution 2: z = 10, so 5y + 180 = 300 from which follows y = 24 and since the total is 100, x = 66.
66 chickens = $198, 24 guinea pigs = $192, 10 geese = $210, total $600.

The next value for z would be 20, but 18*20 is more than 300 so that isn’t possible.

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