Hens, Pigs, & Ducks

Use $S$ for the total Jack has in shillings, and $h,p,d$ for the respective prices of hens, pigs and ducks.

$S=7h+5p$

$S=8p+3d$

$S=4d+9h$

Add the second and third of these equations, and subtract the first:

$S=2h+3p+7d$ giving the answer $\boxed{237}$ .

This is the only order of addition and subtraction that doesn't result in a negative quantity of one of the animals.

Let $S$ be the total Shillings Jack acquired and $x$ be the price of each hen, $y$ be that of each pig, and $z$ be that of each duck.

Then we can set up the Diophantine equations as followed:

$7x + 5y = S$ $8y + 3z = S$ $4z + 9x = S$

We can then rewrite the whole system as matrix multiplication:

$\begin{bmatrix}{7} && {5} && {0} \\ {0} && {8} && {3} \\ {9} && {0} && {4}\end{bmatrix} \begin{bmatrix}{x}\\{y}\\{z}\end{bmatrix} = \begin{bmatrix}{S}\\{S}\\{S}\end{bmatrix} = S\begin{bmatrix}{1}\\{1}\\{1}\end{bmatrix}$

The determinant of the $3\times3$ matrix equals to $7\times 8 \times 4 + 5\times 3\times 9 = 224+135 = 359$

Hence, solving the equation by multiplying the inverse matrix, we will obtain:

$\begin{bmatrix}{x}\\{y}\\{z}\end{bmatrix} = \dfrac{S}{359} \begin{bmatrix}{8\times 4} && {-5\times 4} && {5\times 3} \\ {-9\times -3} && {7\times 4} && {-7\times 3} \\ {8\times -9} && {-5\times -9} && {7\times 8}\end{bmatrix} \begin{bmatrix}{1}\\{1}\\{1}\end{bmatrix} = \dfrac{S}{359}\begin{bmatrix}{32-20+15}\\{27+28-21}\\{-72+45+56}\end{bmatrix} = \dfrac{S}{359}\begin{bmatrix}{27}\\{34}\\{29}\end{bmatrix}$

Since $S$ is prime as stated in the question and so is $359$ , we can conclude that $S = 359$ . Otherwise, $x,y,z$ will not be integers and so contradicted. Thus, $x = 27$ ; $y = 34$ ; and $z = 29$ .

Now let us sum up all the costs: $x + y +z = 27+34+29 = 90$ . Then $359 \equiv 1 \pmod{90}$ , having $1$ Shilling short to make up perfect sets of these animals. Therefore, we need to find the way to put this $-1$ remainder within $x,y,z$ combination.

First, let us rewrite $y = x + 7$ and $z = x + 2$ . Hence, we can rewrite the new Diophantine equation:

$359 = 4x + 4(x+2) +4(x+7) -1 = 12x + 35$

Now we just need to write $35$ in terms of 2- and 7-multiples, for $a,b$ stands for numbers of ducks and pigs purchased: $35 = 2a + 7b$ .

Clearly, $b$ needs to be odd because $35-2a$ is odd, and $b < 5$ because if $b \geq 5$ , $a$ will not be positive integer.

Thus, only $b=1$ or $b = 3$ works. However, if $b =1$ , then $a = 14$ , but $\dfrac{359}{27} < 14$ , meaning Jack can not buy over $13$ animals even by buying all the cheapest hens.

As a result, $b = 3$ is the only solution, leading to $a = 7$ .

That means, Jack would be able to buy $3$ pigs and $7$ ducks, and by simple subtraction, we will calculate $2$ hens as well:

$2x+ 3y + 7z = 2\times 27 + 3\times 34 + 7\times 29 = 359$

Finally, Jack had purchased $2$ hens, $3$ pigs, and $7$ ducks.

Let the number of gold coins, and the prices of hen, pig, and duck be $g$ , $h$ , $p$ , and $d$ respectively. Then we have:

$\begin{cases} 7h + 5p + 0d = g & ...(1) \\ 0h + 8p + 3d = g & ...(2) \\ 9h + 0p + 4d = g & ...(3) \end{cases}$

\(\begin{array} {} 8 \times (1) - 5 \times (2): & 56 h + 0p - 15d = 3g & ...(4) \end{array} \)

From $15 \times (3) + 4 \times (4): \ 359 h = 27g$ . Since $g$ and 359 are primes and $h$ an integer, this means that $g=359$ and $h = 27$ . Substituting the two values in $(1): \implies p = \dfrac {359-7(27)}5 = 34$ and $(3): \implies d = \dfrac {359-9(27)}4 = 29$ .

Let the number of hens, pigs, and ducks purchased be $k$ , $m$ , and $n$ respectively. Then we have:

$\begin{aligned} 27k + 34m + 29n & = 359 \\ \implies 359 - 27k - 29n & \equiv 0 \text{ (mod 34)} \\ 19 + 7k + 5n & \equiv 0 \text{ (mod 34)} \end{aligned}$

Putting $k=1$ $\implies 26 + 5n \equiv 0 \text{ (mod 34)}$ $\implies n \equiv 22$ . But it is unacceptable because $22 > \left \lfloor \dfrac {359}{29} \right \rfloor = 12$ the maximum possible value of $m$ .

Putting $k=2$ $\implies 33+5n \equiv 0 \text{ (mod 34)}$ $\implies n \equiv 7$ acceptable. Implying that $m = \dfrac {359-27(2)-29(7)}{34} = 3$ . Therefore, $\overline{kmn} = \boxed{237}$ .

Let x,y,z be the unit cost of a hen,pig, and duck respectively. Then we have the equations: 7x + 5y = p, 8y + 3z = p, 4z + 9x = p.By eliminating y and z, we get 359x = 27p. But 359 is a prime, so x = 27. Substituting in the other equations, y = 34 and z = 29. If he went home with a hens, b pigs, and c ducks, then 27a + 34b + 29c = 359. Clearly b < 9. Substituting values for b, and finding a,c results in a + 2, b = 3, c = 7.