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During a certain game, after each turn, a player’s points are doubled

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During a certain game, after each turn, a player’s points are doubled  [#permalink]

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New post 21 Aug 2018, 20:54
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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?

(1) Player A takes exactly 4 turns.

(2) Player A’s starting score was not a multiple of 2.

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During a certain game, after each turn, a player’s points are doubled  [#permalink]

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New post 21 Aug 2018, 21:41
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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


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Re: During a certain game, after each turn, a player’s points are doubled  [#permalink]

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New post 21 Aug 2018, 22:18
is there any other way to solve statement b ?
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Re: During a certain game, after each turn, a player’s points are doubled  [#permalink]

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New post 21 Aug 2018, 22:57
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rahulkashyap wrote:
is there any other way to solve statement b ?


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 \(80-x=5...x=80-5=75\)
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Re: During a certain game, after each turn, a player’s points are doubled  [#permalink]

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New post 23 Aug 2018, 23:00
Bunuel wrote:
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?

(1) Player A takes exactly 4 turns.

(2) Player A’s starting score was not a multiple of 2.


Bunuel VeritasKarishma Pls help to solve statement B.
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Re: During a certain game, after each turn, a player’s points are doubled  [#permalink]

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New post 25 Aug 2018, 05:24
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siddreal wrote:
Bunuel wrote:
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?

(1) Player A takes exactly 4 turns.

(2) Player A’s starting score was not a multiple of 2.


Bunuel VeritasKarishma Pls help to solve statement B.



When the player completes the game (reaches a score of 0), we know what if his number of points before that were N,

2N - 80 = 0
N = 40

Now, can we say what the score was in the previous turn? Sure. We got 40 by doubling the previous score and subtracting 80.

2M - 80 = 40
M = 60

Now, can we say what the score was in the previous turn? Sure. We got 60 by doubling the previous score and subtracting 80.

2L - 80 = 60
L = 70

Now, can we say what the score was in the previous turn? Sure. We got 70 by doubling the previous score and subtracting 80.

2K - 80 = 70
K = 75
(not an even score)

Now, can we say what the score was in the previous turn? Sure. We got 75 by doubling the previous score and subtracting 80.

2J - 80 = 75
J = 77.5

The reason we did not need to do all this calculation backwards was this - Every time, we are adding 80 to the current score and dividing by 2. As long as our current score is even,
(Even + 80) = Even, so on dividing by 2 we get an integer.
The moment out current score becomes odd (75 in our case), the previous score will not be an integer because Odd + 80 = Odd. So on dividing by 2, we get a decimal. But every score must be an integer. This tells us that there is only one odd score that anyone can have and that is in the beginning of the game and it is 75 only (as calculated above). But since this is a DS question, we didn't really need to do the calculation as long as we know that we will get a single unique value of the starting score.
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Re: During a certain game, after each turn, a player’s points are doubled  [#permalink]

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Re: During a certain game, after each turn, a player’s points are doubled   [#permalink] 11 Sep 2019, 11:11
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