100 animals cost 4000. calves, 120 each; lambs, 50 each; piglets, 25. How many of each did he buy?Numer theory

Let the number of calves be a, the number of lambs be b
and the nuber of piglets be c.

The total number of heads
is:

a + b + c = 100

The price
paid to buy 100 heads of animals is 4000 monetary units, such
as.

120a + 50b + 25c =
4000

We'll use the first equation to write c in terms of a
and b.

c = 100 - a - b

We'll
substitute c into the 2nd equation:

120a + 50b + 25(100 - a
- b) = 4000

120a + 50b + 2500 - 25a - 25b =
4000

We'll combine like
terms:

95a + 25b = 1500

We'll
divide by 5:

19a + 5b = 300

We
notice that if a = 0 and b = 60, then (0,60) represents a solution of this
equation.

We'll consider as solution of the equation a'= 5x
and b' = 60 - 19x, where x is an integer number.

The values
of a,b,c must be natural numbers since they represent the number of
animals.

For b' to be positive, x must be positive but
smaller, or equal to 3.

Therefore,
we'll have:

For t = 1 => a' = 5 and b' = 60 - 19 =
41 and c = 100 - 5 - 41 = 54

For t = 2 => a' = 10;
b' = 60 - 38 = 22 => c = 100 - 10 - 22 = 68

For t =
3 => a' = 15 ; b' = 60 - 57 = 3 => c = 100 - 15 - 3 =
82.

Therefore, there are 3 possibilities: 5
calves ; 41 lambs and 54 piglets or 10 calves, 22 lambs and 68 piglets or 15 calves, 3
lambs and 82 piglets.