Three people play a game in which there are always two winners and one loser. Th

Question

Three people play a game in which there are always two winners and one loser. They have the understanding that the loser gives each winner an amount equal to what the winner already has. After three games, each lost just once and each has $24. With how much money did each begin? Three people play a game in which there are always two winners and one loser. They have the understanding that the loser gives each winner an amount equal to what the winner already has. After three games, each lost just once and each has $24. With how much money did each begin?

Explanation / Answer

Let us suppose the three players to be a,b and c

Let us assume each one started with x,y and z dollars

After first round let us assume a lost (as per their understanding),then

a has x-y-z

b has 2y and

c has 2z

similarly after second round if b lost then

a has 2(x-y-z)

b has 2y-(x-y-z)-2z and

c has 4z

after the third round if c lost,then

a has 4(x-y-z)

b has 2(2y-(x-y-z))

c has 4z-2(x-y-z)-(2y-(x-y-z))

Solving these equations using Guass Jordan elimination method

a has $39

b has $21 and

c has $12

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