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

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### 1 Answer

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.**