3 Little Pigs Riddle

Firstly, organize the information as below:

To pay the least possible amount of wages, the wolf should employ pigs B and C as long as possible. However, as both pigs have to rest for at least a day, the work will be done by pigs A & C and A & B on the respective days. Therefore, we have:

Solving Equation:

$\begin{array} { c } { n \left( \frac { 1 } { 12 } + \frac { 1 } { 16 } \right) + \left( \frac { 1 } { 9 } + \frac { 1 } { 12 } \right) + \left( \frac { 1 } { 9 } + \frac { 1 } { 16 } \right) = 1 } \\ { \mathrm { n } = \frac { 13 } { 3 } = 4.333 \dots } \end{array}$

Put n=4, we find that the total work done= $\frac{137}{144}$ , with $\frac{7}{144}$ remaining.

To avoid being charged another full day's wage (which is $250+$150=$400 at least, larger than any pig's individual daily wage), we may substitute pig B or C by pig A for 1 day.

$\begin{array} { l } { \text { Daily work done by Pig A is more than that of Pig B by } \frac { 1 } { 9 } - \frac { 1 } { 12 } = \frac { 1 } { 36 } } \\ { \text { Daily work done by Pig A is more than that of } \operatorname { Pig } \mathrm { C } \text { by } \frac { 1 } { 9 } - \frac { 1 } { 16 } = \frac { 7 } { 144 } } \end{array}$

Therefore, if Pig C is substituted by Pig A for 1 day, the work can be done in 6 complete days, while keeping the wage as low as possible.

The arrangement is therefore:

The least possible amount of wage= $350 \times 3 + 250 \times 5 + 150 \times 4 = ( \$ ) 2900$

Let A be the eldest pig, B be the middle pig, and C be the youngest pig. Then for one day:

A & B builds $\frac{28}{144}$ of the house for $600

A & C builds $\frac{25}{144}$ of the house for $500

B & C builds $\frac{21}{144}$ of the house for $400

By the pigs' conditions, each pig must have at least 1 day off and 2 pigs must work on the house each day, so there must be at least 1 day of A & B, 1 day of A & C, and 1 day of B & C, for a total of $\frac{74}{144}$ of the house built at $1500.

This leaves $\frac{70}{144}$ of the house remaining, which can be done minimally in 3 more days by either 2 more days of B & C and 1 day of A & B, or by 2 more days of A & C and 1 more day of B & C, both scenarios costing the wolf $1400 more. Either way, the total cost is $1500 + $1400 = $2900 .

After the house was built, the wolf discovered that two of the pigs substituted unauthorized straw and sticks for bricks, and confronted each of them at their private residences rather huffily ...

Let $x$ be the number of days pigs $A, B$ work together, $y$ be the number of days pigs $A, C$ , and $z$ be the number of days pigs $B, C$ , respectively.

Since the pigs need a day off some time in this period, $x,y,z$ can't be 0. Otherwise, such pig would work throughout the whole construction length.

In order to avoid the full-day charge, we have to calculate the way this job is complete within full total days, and the work portions done in each day will equal to $\dfrac{1}{9}$ , $\dfrac{1}{12}$ , and $\dfrac{1}{16}$ for pigs $A, B, C$ respectively. That is, $x,y,z$ will be non-negative integers as in this Diophantine equation:

$(\dfrac{1}{9} + \dfrac{1}{12})x + (\dfrac{1}{9} + \dfrac{1}{16})y + (\dfrac{1}{12} + \dfrac{1}{16})z = 1$

Multiplying by the least common multiple of $144$ , we will obtain:

$(16+12)x + (16+9)y + (12+9)z = 144$

$21(x+y+z) + 7x + 4y = 144$

Since $144 = 21\times 6 + 18$ , we can conclude that $6 = x+y+z$ . Then we will have to rewrite the remainder $18$ as a Diophantine equation of $18 = 7x + 4y$ . Because $18$ is not a multiple of $7$ or $4$ , $x,y$ must be positive integers, and clearly with a little force, there is only one way to write $18 = 2\times 7 + 1\times 4$ .

Thus, $x = 2$ and $y=1$ , leading to $z=3$ .

Then for each day $x$ , it will cost $350+250 = \$600$ .

For each day $y$ , it will cost $350+150 = \$500$ .

For each day $z$ , it will cost $250+150 = \$400$ .

Finally, the total payment will be: $2\times 600 + 1\times 500 + 3\times 400 = \$2,900$ .

Pig $A$ costs $\$3150$ per unit house built, Pig $B$ costs $\$3000$ per unit house built, and Pig $C$ costs $\$2400$ per unit house built. To minimise cost, we want to Maximise the use of Pigs $B$ & $C$ , and minimise the use of Pig $A$ .

If we only used Pigs $B$ & $C$ every day, they would build a house within $7$ days, costing $\$2800$ . However, as each Pig needs one day off, the cheapest we could do it in $7$ days is $2$ , $6$ & $6$ days working between Pigs $A$ , $B$ & $C$ respectively, costing $\$3100$ .

Next, we try to reduce the number of days:

Let's assume it can be done in $6$ days, with a baseline of $2$ , $5$ & $5$ days working, with a cost of $\$2700$ . However, this does not quite complete a house in that time.

The cheapest way to build faster, would be by swapping either $B$ for $A$ or $C$ for $B$ for $1$ day, costing an extra $\$100$ . However, only swapping a single day does not save enough time to complete the house in $6$ days: for that, we need $2$ swaps, costing us $\$200$ extra, resulting in a total cost of $\boxed{\$2900}$ .

We cannot build a house in fewer days than this, as the fastest option of $4$ , $4$ & $2$ is only able to build ~ $\frac{9}{10}$ of a house in that time.

The pigs need to work in pairs, their productivity is:

Since each team must be active for at least a day, this accounts for building $\frac{74}{144}$ of a  house. Note that team C offer the most value for money, then team B, then team A. The cheapest way to realise the other $\frac{70}{144}$ of the house is however to hire team B for two more days, and team C for 1 more day, that way they could produce $\frac{21+25+25}{144}=\frac{71}{144}$ of a house.

So together he hires team A for 1 day, team B for 3 days and team C for 2 days. The cost will be $1×600+3×500+2×400=\$2900$ .

Let

(1) x be the number of days for the pair (1,2), the amount of completed work=x(1/9)+x(1/12)=x(7/36),

(2) y be the number of days for the pair (1,3), the amount of completed work=y(1/9)+y(1/16)=y(25/144),

(3) z be the number of days for the pair (2,3), the amount of completed work=z(1/12)+z(1/16)=z(7/48).

Write the total work =1 as the sum of above three pair of works:

x(7/36)+y(25/144)+z(7/48)=1.

Solve the above Diophantine equation using WolframAlpha to get solution:

x=2, y=1 and z=3 and the total cost as:

2 (350+250)+(350+150)+3 (250+150)

=2900

Answer=2900