[madadi]

Originally Posted by phasmophobe007

There are 500 balls.A person can pick a minimum of 5 balls and maximum of 10 balls.A & B are playing this game.A starts the game.How many balls A should pick to win if:
i)Last picker wins the game
ii)Last picker looses the game.


I don't have the key to above question.
puys,pls post ur various approaches.

i feel the answer to both these parts is the same.... which is 5...
explanation:
1) A starts n has to be the one who picks up the ball last, to be the winner... now since, a minimum of 5 balls are to be picked essentially,
the series wud go as--->
A B A B A B A B A
hence A picks up 5 balls, n emerges as the winner...

2) A starts again... the last picker wud lose, in this case... the series, hence formed wud be--->
A B A B A B A B A B
B picks up the ball last, hence loses... wich implies A is the winner.. but the no of balls picked by A wud remain 5 itself...

correct me if i'm wrong, puys..
p.s: shudn't the question have been framed like... find the minimum no of balls A must pick up in order to win??

[themystique]

NUMBER OF CUBES IN 27000 :

cubes of 1,2,3,5,6,10,15,30 are present.

hence 8.
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[themystique]

Originally Posted by phasmophobe007

There are 37 balls in a bag.A person can pick minimum of 1 ball and maximum of 3 balls from this bag.The person who picks the last ball wins the game.A & B are playing this game,starting with A and alternatively picking the ball.If A wants to win the game ,how many balls shall A pick?do post ur approach also


Answer to aforesaid is
A shall pick 1 ball to win.
Approach-The approach given by srikar is a generalised one for sure.
But my approach-
ball no. 37 shall be picked up by A
so A should pick ball no. 33,29,25,.....,5,1
Hence A should pick 1 ball to win.

for this question points to be kept in mind-
A picks the first ball.After this picks(1 to 3) no. of balls.
How many balls A will pick now?
The total no. of balls picked by B(in 2nd turn) and by A(3rd turn) shall be equal to 4.If B pick 2 balls in his 2nd turn then A should also pick 2 and so on...

General formulae for above type of question
No. of ball A shall pick=Total no.of balls%(Minimum+maximum no. of balls picked)

@phasmo,

could you please tell me if the formula that you provided is true in all cases?
and in which cases it could be applied and in which it shouldn't be?

thanks </div>

[ravijanjanam]

Originally Posted by 80199

In the X-Y plane, the area of the region bounded by the graph of |x + y| + |x -y| - 4 is

Choose one answer.
a. 8 b. 12 c. 16 d. 20

sol: |x| can be expressed as +/- x

therfore |x+y|= +/-(x+y)
|x-y|= +/-(x-y)

=> case1: +(x+y)+(x-y)-4 = Z(suppose)

=> 2x-4=Z------------eq 1
=> case 2: -(x+y)-(x-y)-4 =Z1 (suppose)

=> -2x-4=Z1-----------eq 2

plotting eqns 1 and 2 on graph provieds us triangle with points (-2,0),(0,-4),(2,0)

therfore finding the area of this triangle is the solution

[Udit Mohan]

hi folks ,
pls solve these questions ...
*** How many terms are there in the expression : (y+y^3+y^5+y^7)^5
*** Find the no of positive integral solution of A*B*C*D*E=1050??
***50 kids are standing in a cirular arrangement and one of them is having a ball .....the ball is passed counter-clock wise skipping 0,1,2,3,4 & so on kids respectively i.e. the 1st kid gives it to 2 nd kid, then 2nd kid gives it to 4th kid, 4th to 7 th & so on .....after how many hands change will the ball back to the first kid ???

[dagaritesh]

***50 kids are standing in a cirular arrangement and one of them is having a ball .....the ball is passed counter-clock wise skipping 0,1,2,3,4 & so on kids respectively i.e. the 1st kid gives it to 2 nd kid, then 2nd kid gives it to 4th kid, 4th to 7 th & so on .....after how many hands change will the ball back to the first kid ???[/quote]

Is it 24.

By my understanding n*(n+1)/2 should be a multiple of 50 and n should be minimum.

Hence 24?

[ashishkrsaini]

Originally Posted by prince_chittu

@Srikar

Can u plz elaborate on the answers u hv given of the this question
There are 500 balls.A person can pick a minimum of 5 balls and maximum of 10 balls.A & B are playing this game.A starts the game.How many balls A should pick to win if:
i)Last picker wins the game
ii)Last picker looses the game.

Okay Here is my brain work ..
A & B will always make a combination of 15 score if B picks first and A follows ie BA = 15. eg B = 5 , A=10 or B=10, A=5

Now In case I, let A pick x balls. so we have x + 15N = 500
At N=33 and x = 5 we have a solution i.e if A picks 5 Balls initially, A will always pick last ball. Sequence would be A BA BA BA BA ...

Now In case II, let A pick x balls in first turn. Assuming B will have to pick last set of balls. In this case, B should be left with 5 to 9 balls in the last turn. P.S - In case 9 balls are left, B cannot pick 5 cause this will make 4 balls left, which cannot be picked by A!

The sequence would be A BA BA BA BA B
The equation would be x + 15N + (5...9) = 500
i.e x + 15N = (491...495)
Now for N = 32, x = 11...15
N = 33 x = 0....5

So that means, If A starts, A cannot Win.

Others .. please put your thoughts on this.

[dagaritesh]

[quote=Udit Mohan;1165346]hi folks ,
pls solve these questions ...
*** How many terms are there in the expression : (y+y^3+y^5+y^

Are there 24 terms?

[themystique]

Originally Posted by dagaritesh

***50 kids are standing in a cirular arrangement and one of them is having a ball .....the ball is passed counter-clock wise skipping 0,1,2,3,4 & so on kids respectively i.e. the 1st kid gives it to 2 nd kid, then 2nd kid gives it to 4th kid, 4th to 7 th & so on .....after how many hands change will the ball back to the first kid ???

Is it 24.

By my understanding n*(n+1)/2 should be a multiple of 50 and n should be minimum.

Hence 24?[/quote]

yes, i believe this is right.
i took the cue from you and tried it out for ten members.
it is the smallest value of n(n+1)/2 to be a multiple o the given no. of people.

hence 24 is correct

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[themystique]

@ udit :
*** Find the no of positive integral solution of A*B*C*D*E=1050??

is there any condition that a,b,c,d,e are not distinct?

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