During a certain game, after each turn, a player’s points are doubled and then reduced by 80. Players can only have a whole number of points and take turns until their score reaches exactly zero. Assuming no other points are gained or earned, how many points did player A start with?

Let the initial point be x, so the points after one turn \(2x-80=2^1*x-80*1=2^1x-80*(2^1-1)\) and the next \(2(2x-80)-80=4x-240=2^2x-80*3=2^2x-80*(2^2-1)\)and so, the nth turn \(= 2^nx-80*(2^n-1)\)(1)

Player A takes exactly 4 turns.so after 4th turn points = 0..

so two ways

a) 4th turn - \(2^4x-80*(2^4-1)=0.........16x-80*15=0......16x=80*15......x=5*15=75\)

b) work backwards - after 4th turn - 0, so before 4th turn - 0+80/2=40, before 3rd turn - (40+80)/2=60, before 2nd turn - (60+80)/2=70, before first turn - (70+80)/2=75

sufficient

(2)

Player A’s starting score was not a multiple of 2.A straight logical answer is YES..since we have a term being multiplied by 2, there will be only one term which will be a non-multiple of 2, because a term prior to it will be a decimal.

other way..nth term = \(2^nx-80*(2^n-1)=0........2^n(80-x)=80\)

x is an ODD integer, and 80-x has to be a multiple of 5, so 80-x can be 5, 15 , 25 ....

Only \(2^n*5 = 80\) when n is 4, so \(80-x=5...x=80-5=75\)

work backwards - after last turn - 0,

so before that turn - \(\frac{0+80}{2}=40\),

before that turn - \(\frac{(40+80)}{2}=60\),

before that turn - \(\frac{(60+80)}{2}=70\),

before that turn - \(\frac{(70+80)}{2}=75\)

before that - \(\frac{(75+80)}{2}=77.5\)... but this is not a whole number

so he started with 75

sufficient

D

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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