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#1 2011-06-05 04:25:24

wintersolstice
Real Member
Registered: 2009-06-06
Posts: 114

"Dropping Balls" Puzzle

One of my obsessions with puzzles is to extend them and make harder versions

like this puzzle from the site:

Dropping BallsImagine that you had 3 balls and 1000 storey building

What would be the solution now?

(Everything else is the same.)

Last edited by wintersolstice (2011-06-05 04:25:43)


Why did the chicken cross the Mobius Band?
To get to the other ...um...!!!

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#2 2011-06-05 04:48:09

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,673

Re: "Dropping Balls" Puzzle

hi ws


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#3 2011-06-05 10:16:00

wintersolstice
Real Member
Registered: 2009-06-06
Posts: 114

Re: "Dropping Balls" Puzzle

No thats when you have 2 balls. With 3 you can do it in fewer!


Why did the chicken cross the Mobius Band?
To get to the other ...um...!!!

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#4 2011-06-05 17:45:33

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,555

Re: "Dropping Balls" Puzzle

I could add this to the puzzles if you wish.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#5 2011-06-05 23:23:37

ZHero
Real Member
Registered: 2008-06-08
Posts: 1,889

Re: "Dropping Balls" Puzzle

@MathsIsFun: The puzzle at MathIsFun reads "What is the maximum number of times you have to drop the snooker balls.....". Well, I guess, I can drop the balls a maximum of 100 times starting from 1st floor continuing upto 100th floor? dunno


If two or more thoughts intersect with each other, then there has to be a point.

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#6 2011-06-06 01:54:26

gAr
Member
Registered: 2011-01-09
Posts: 3,479

Re: "Dropping Balls" Puzzle

Hi wintersolstice,


"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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#7 2011-06-06 03:33:02

wintersolstice
Real Member
Registered: 2009-06-06
Posts: 114

Re: "Dropping Balls" Puzzle

gAr wrote:

Hi wintersolstice,

no but very close:D


Why did the chicken cross the Mobius Band?
To get to the other ...um...!!!

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#8 2011-06-06 05:27:19

gAr
Member
Registered: 2011-01-09
Posts: 3,479

Re: "Dropping Balls" Puzzle

Hi wintersolstice,


"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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#9 2011-06-06 10:01:07

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,673

Re: "Dropping Balls" Puzzle

hi ws

Last edited by anonimnystefy (2011-06-06 10:01:18)


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#10 2011-06-06 15:10:05

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,555

Re: "Dropping Balls" Puzzle

ZHero wrote:

"What is the maximum number of times you have to drop the snooker balls....

Good point. Changed it to "least". Thanks.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#11 2011-11-12 04:43:31

wintersolstice
Real Member
Registered: 2009-06-06
Posts: 114

Re: "Dropping Balls" Puzzle

I still need a procedure but that's the answer:D


Why did the chicken cross the Mobius Band?
To get to the other ...um...!!!

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#12 2012-01-14 19:33:08

sny1
Member
Registered: 2012-01-14
Posts: 1

Re: "Dropping Balls" Puzzle

In case of "Droppng Balls Puzzle", the minimum no. of cases of dropping balls should be 12 if it doesn't break until 99th floor......
Wat u say ?????.

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#13 2012-09-30 17:11:15

vikramhegde
Member
Registered: 2012-09-30
Posts: 1

Re: "Dropping Balls" Puzzle

Ok so the procedure is like this.
First consider a triangular number sequence [ fn = (n+1).n/2] for n = 1,2,3..
The 18th number on this sequence is 171.
And sum of all triangular numbers till 171 (the tetrahedral number) crosses 1000
Now, for each subsequent drop of the first ball (when it doesn't break) we add the previous number on the triangular sequence.
So ball number one is dropped for the first time from Floor number 171, second at (171+153)=324, third at (324+136)=460, fourth at (460+120)=580, fifth at (580+105)=685, sixth at (685+91)=776, seventh at 854, eighth at 920, ninth 975, tenth at 999. Let us call these numbers M#1, M#2, M#3....

Now, if it breaks in the first try at 171, we start dropping the ball at floor numbers 18, 18+17, 18+17+16,
Similarly for any break of the first ball on floor M#x, the floor number at which we drop the second ball is given by -
{(M#x-1)+P} and if the second ball doesn't break at this, we continue on the sequence - {(M#x-1)+P+(P-1)}, {(M#x-1)+P+(P-2)},  {(M#x-1)+P+(P-3)}... {(M#x-1)+P+(P-P)} Where P is the position of {M#x - (M#x-1)} on the triangular sequence.

Having broken the second ball somewhere, we go back to the last try where we didn't break it and work our way up with tries on each floor till it breaks.

We will find that 19 is the least number of tries required.

I'm sorry I'm not trained in mathematics and hence have to put it in such a round about manner. I'm not so familiar with the notation and the use of sigma functions and had to invent some notation of my own. I hope I've explained it adequately.

Also, I'd be happy if someone could explain it a more simple manner.

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#14 2012-09-30 18:55:13

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,536

Re: "Dropping Balls" Puzzle

hi vikramhegde

Thanks for the method.  Inventing notation is perfectly acceptable as long as you define your terms (as you have).  And you have arrived at the OP's answer!  So it would seem you are more of a mathematician than you think!  Lot's of us here are 'self-taught'; I could argue the case that makes you a better mathematician!

Welcome to the forum!  smile

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#15 2012-10-06 06:39:28

wintersolstice
Real Member
Registered: 2009-06-06
Posts: 114

Re: "Dropping Balls" Puzzle

vikramhegde wrote:

Ok so the procedure is like this.
First consider a triangular number sequence [ fn = (n+1).n/2] for n = 1,2,3..
The 18th number on this sequence is 171.
And sum of all triangular numbers till 171 (the tetrahedral number) crosses 1000
Now, for each subsequent drop of the first ball (when it doesn't break) we add the previous number on the triangular sequence.
So ball number one is dropped for the first time from Floor number 171, second at (171+153)=324, third at (324+136)=460, fourth at (460+120)=580, fifth at (580+105)=685, sixth at (685+91)=776, seventh at 854, eighth at 920, ninth 975, tenth at 999. Let us call these numbers M#1, M#2, M#3....

Now, if it breaks in the first try at 171, we start dropping the ball at floor numbers 18, 18+17, 18+17+16,
Similarly for any break of the first ball on floor M#x, the floor number at which we drop the second ball is given by -
{(M#x-1)+P} and if the second ball doesn't break at this, we continue on the sequence - {(M#x-1)+P+(P-1)}, {(M#x-1)+P+(P-2)},  {(M#x-1)+P+(P-3)}... {(M#x-1)+P+(P-P)} Where P is the position of {M#x - (M#x-1)} on the triangular sequence.

Having broken the second ball somewhere, we go back to the last try where we didn't break it and work our way up with tries on each floor till it breaks.

We will find that 19 is the least number of tries required.

I'm sorry I'm not trained in mathematics and hence have to put it in such a round about manner. I'm not so familiar with the notation and the use of sigma functions and had to invent some notation of my own. I hope I've explained it adequately.

Also, I'd be happy if someone could explain it a more simple manner.

just to correct you on that it should be M#x-1 + p + 1 to see why imangine dropping the first ball from 172. You can still do it even if it breaks on that floor other that your solution is spot on:D

Last edited by wintersolstice (2012-10-06 06:40:56)


Why did the chicken cross the Mobius Band?
To get to the other ...um...!!!

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#16 2012-10-22 04:41:23

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,758

Re: "Dropping Balls" Puzzle

Hi vikramhegde;

vikramhegde wrote:

We will find that 19 is the least number of tries required.

Have you gone to 10000 floors? I am getting 33 for that.


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

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