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

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

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

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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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

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

@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?

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

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

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

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

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

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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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

ZHero wrote:

"What is the

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

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

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

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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

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!

Bob

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

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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