Puzzles with numbers and things

[snookersfun]

I think, that is good, at least it matches my solution. But feel free to come up with a different constellation or the three dimensional one.
second part of the question? although trivial

[davis_greatest]

Second part - for n people / apes, the greatest number of balls anyone can receive is min(n-1 , 5).

i.e. answer won't change when adding new members

[snookersfun]

So far, I would say all good for the point. But feel free to continue....
Are you updating the scoreboard then, including the previous round (which definitely looks solved to me).

[The Statman]

Yep, the previous solution is solved - you can update the scoreboard after this round is completed.

[davis_greatest]

Yep, the previous solution is solved - you can update the scoreboard after this round is completed.

How do you want to score that birthday question round - a point each for Semih and me?

[davis_greatest]

HERE IS THE SCOREBOARD AFTER ROUND 55

snookersfun……………………….…..21½
abextra……………………………..…...11
Vidas……………………………………….10½
davis_greatest…………………..……8
robert602…………………………………6
elvaago...............................4
The Statman……………………..……3
Semih_Sayginer.....................2½

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[Snooker Rocks!]

Can I give a little problem?

WITHOUT the use of a calculator:

(1+1/2)(1+1/3)(1+1/4).......(1+1/2000)(1+1/2001)

How can you solve this, using a neat, and simple method?


p.s. I didn't make this question up; I took a UK Maths Challenge paper, and saw this question and found a neat solution

[davis_greatest]

Welcome to this thread, Snooker Rocks! I'll send you my answer by PM, so that others can have a go but so that I'll still be eligible for the point!

I'll also try to think of a question in a few minutes, to run concurrently...

[davis_greatest]

Round 56; and Round 57 - "Stand in line"

Round 56: Snooker Rocks!'s question above - I'll share the point for that one with the first person to answer it on this thread!

================================================== =

Round 57: Stand in line!

We're still playing games at Oliver's birthday party. This is the party of all parties! This time, we all stand in a straight line, holding hands, with Oliver on the far left and Gordon on the far right, and abextra with her hands gripped by two excited apes.

In fact, the order is:

Oliver----austrian_girl----davis_greatest----April Madness----Robert602----snookersfun----Semih_Sayginer----The Statman----Vidas----elvaago----Obligation----Snooker Rocks!----Charlie----abextra----Gordon


(apologies if any contributors to this thread have been omitted and are not playing this game at the party - if so, you can please referee this one)


At any time, anyone who has two or more people/apes on his or her side, can move two places to that side.

For example, The Statman might decide to move two places to his right and stand between elvaago and Obligation; or instead two places to his left and stand between Robert602 and snookersfun.

As another example, at the outset, austrian_girl could move two places to her right and stand between April Madness and Robert602, but she could not initially move to her left because she only has one person / ape on her left at the start (and so cannot move two places).


We can all move at any time, as often as we wish. Our aim in this game is for us all to be standing in the exact reverse order of how we started. So, we need to finish with Gordon on the far left, Oliver on the far right, and everyone else in between exactly reversed.


Your question is:
Is this possible? If yes, explain how we achieve it. If no, explain why not.

[elvaago]

I'll rewrite this as:
3/2 * 4/3 * 5/4 * 6/5 * ... * 2002/2001

This will look like:
(3 * 4 * 5 * 6 * ... * 2002)
----------------------------
(2 * 3 * 4 * 5 * ... * 2001)

You can cross out the 3 until the 2001 below and above the line. So you end up with:
2002/2 = 1001

[davis_greatest]

I'll rewrite this as:
3/2 * 4/3 * 5/4 * 6/5 * ... * 2002/2001

This will look like:
(3 * 4 * 5 * 6 * ... * 2002)
----------------------------
(2 * 3 * 4 * 5 * ... * 2001)

You can cross out the 3 until the 2001 below and above the line. So you end up with:
2002/2 = 1001

I'll give davis_greatest and you half a point each as davis_greatest sent the same answer to Snooker Rocks! by private message and I think a whole point each would be a bit generous...

SO HERE IS THE SCOREBOARD AFTER ROUND 56

snookersfun……………………….…..21½
abextra……………………………..…...11
Vidas……………………………………….10½
davis_greatest…………………..……8½
robert602…………………………………6
elvaago...............................4½
The Statman……………………..……3
Semih_Sayginer.....................2½

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)


Round 57 is awaiting a solution

[elvaago]

Oh I don't care about points all that much. For all I care, you get the whole point for being first. I'm just happy I solved one for a change. :-D

[davis_greatest]

Oh I don't care about points all that much. For all I care, you get the whole point for being first. I'm just happy I solved one for a change. :-D

elvaago, these points are highly sought after, so you should be proud. There is even the possibility that, once 100 points are reached, these might be exchangeable for a banana.

[elvaago]

In that case, I WANT A WHOLE POINT!!!!111oneoneeleven

[snookersfun]

Round 57: Stand in line!

We're still playing games at Oliver's birthday party. This is the party of all parties! This time, we all stand in a straight line, holding hands, with Oliver on the far left and Gordon on the far right, and abextra with her hands gripped by two excited apes.

In fact, the order is:

Oliver----austrian_girl----davis_greatest----April Madness----Robert602----snookersfun----Semih_Sayginer----The Statman----Vidas----elvaago----Obligation----Snooker Rocks!----Charlie----abextra----Gordon

At any time, anyone who has two or more people/apes on his or her side, can move two places to that side.
We can all move at any time, as often as we wish. Our aim in this game is for us all to be standing in the exact reverse order of how we started. So, we need to finish with Gordon on the far left, Oliver on the far right, and everyone else in between exactly reversed.

Your question is:
Is this possible? If yes, explain how we achieve it. If no, explain why not.

Reversing the order of the line-up is not possible for 15 people by moving over two places at a time.
One can show though, that it is possible to transpose pairs succesfully by leapfrogging in steps of 7. One can basically start at one end (1 or 15) and leapfrog this number into the 'new' middle, followed by the next number at the other side and so on alternating sides (1,15,2,14....8). We need 7x15 such steps = odd number. Therefore the inversion can't be done using jumps by 2.