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

@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?

gAr wrote:

Hi wintersolstice,

no but very close:D

hi ws

Last edited by anonimnystefy (2011-06-07 08:01:18)

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.

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 (2012-10-07 05:40:56)