Kong Off 119

[Weehawk]

It is the year 2062. Richie Knucklez IV is hosting Kong Off 119. RK4 has every known remaining Donkey Kong machine available for use in the competition.

On Day 1, he has the same number of competitors assigned to each machine.

On Day 2, ten of the machines have broken down and as a result, each remaining machine has one additional competitor assigned to it.

On Day 3, fifteen more of the machines fail, leaving 3 more competitors per remaining machine than there had been on Day 1.

How many people (technically one of them was an android with the late Dean Saglio's memory engrams stored in its neural network, but disregard that) competed in Kong Off 119?

For the first person to post the correct answer with explanation, I have a DK Forum t-shirt: (if you already have one, you can designate another forum member to receive it, or sell it on the black market, I don't care)

[VON]

900 players.  100 machines.

Day 1 starts with 9 players on each machine.  10 machines go down on Day 2, adding 90 players to the remaining 90 machines; now 10 players on each machine.  15 additional machines break down on Day 3, adding 150 players to the remaining 75 machines; now 12 per machine. 

[Weehawk]

This was an easy one. And it is just a rewording of Sam Loyd's "Great Picnic" puzzle.

You have 3 variables:

c is the number of competitors.

p is the number of people per machine on Day 1.

m is the number of machines.

the puzzle gives you three independent equations using those variables:

pm = c

(p+1)(m-10) = c

(p+3)(m-25) = c

For those not familiar with (or who have forgotten) the use of matrices and determinants to solve linear systems, simple algebraic substitution will suffice.

Substituting pm for c in the second and third equations gives us:

-10p + m -10 = 0

-25p +3m -75 = 0

Solving either one for p or m and substituting into the other gives p=9 and m = 100

c = pm = 9*100 = 900

QED

Ross, PM me your address and size.