Puzzles with numbers and things

[snookersfun]

Yes I agree Monique, I thought I had a 41 for this arrangement but alas 40 seems to be the most.

had lot of dodgy 41s, but think I do have a good one now...

and if we can do 'not strictly 2 dimensional stuff', 8524

[moglet]

and if we can do 'not strictly 2 dimensional stuff', 8524

I can squeeze two more, 8526, for this arrangement

[snookersfun]

I can squeeze two more, 8526, for this arrangement

gaaaaahhh, me too, can't think straight though

[abextra]

had lot of dodgy 41s, but think I do have a good one now...

and if we can do 'not strictly 2 dimensional stuff', 8524

I can squeeze two more, 8526, for this arrangement

I maybe have 42 on the floor, probably because of my good adding skills.

Moving to the third dimension now.

[abextra]

8526??? HOW on earth have you got this many?

[davis_greatest]

Do we need to use all cues?

Yes.

Please would everyone put up his / her arrangement(s). Not sure who is going to count the 8526 though...

And would someone please put up an answer & explanation to round 385.

[Monique]

R385

Oliver takes 49 tables randomly from the West wall, turns them and pushs them to the East wall.

If say n of these 49 tables were facing east, that leaves us with 147-n east facing tables on the West side and adds 49-n to the 98 we have on the East side.

[snookersfun]

I maybe have 42 on the floor, probably because of my good adding skills.

so one of my 41s quick, before abextra is bettering that:
squares-41-2.bmp

55 squares-14 cues

8526??? HOW on earth have you got this many?

will leave that for moglet to explain and count

[abextra]

so one of my 41s quick, before abextra is bettering that:

55 squares-14 cues

Thank you, Snookersfun! Can't better that, lol, it was my adding again, I have only 41 too.

[moglet]

That's a relief abextra, I found the 41s more easily today, so I wondered if you had found a 42 the others had missed.

The glut of bananas, all 8526 of them, was an arrangement of 30 cues laid side by side but slightly apart with the remaining 30 cues laid at right angles over the others, not exactly "on the floor"
Too many squares to count but the total number is given by:

n(n+1)(2n+1)/6

where n= the number of the smallest squares on the side of the grid formed by the "arrangement" - so n=29 total 8555, to remove every "square" we just need to remove all but 1 cue from the top layer, 8555-29=8526.

[davis_greatest]

Oliver takes 49 tables randomly from the West wall, turns them and pushs them to the East wall.

If say n of these 49 tables were facing east, that leaves us with 147-n east facing tables on the West side and adds 49-n to the 98 we have on the East side.

Yes, indeed!

Congratulations also to all who got the 41 or 8526 bananas in the last round. I couldn't better those - as the 8526 does not strictly allow the cues to be laid on the floor, I will accept both the 41 and 8526 answers as the "best possible". So well done!

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

Round 387 - Going Ape in Pentrow

At Barry The Baboon's Birthday party today at his Ball Shop, the apes (all slightly different heights) gathered and played with the balls there. Each ape laid some balls in a square formation - one separate square per ape.

Oliver, being the smallest ape present, laid out a square of 10731 by 10731 balls. Each successively taller ape laid out a square slightly bigger than his predecessor's, with one extra row and column.

There were only baboons and orang-utans at the party (Charlie and Gordon being busy in Pentrow playing in the Apewash Open) and, had Oliver not made it, there would have been equal numbers of each species. All the baboons were taller than the orang-utans, but in total each species had laid out the same number of balls.

How many apes were at Barry's party?

What do you get if you multiply the number of apes by the number of baboons?

Answers by Private Message initially please

[davis_greatest]

Correct answers received to round 387 from snookersfun, moglet and Monique - congratulations! Let's have answers on the thread please (and see whose is the most straightforward)

Round 388 - Crossing the line

Can you draw 9 lines (which may be 'curved' lines) on a snooker table, so that:
- each line touches two spots: at one end must be a baulk colour spot (yellow, green or brown) and at the other end a high value colour spot (blue, pink, black);
- each of the baulk spots is paired once with each of the high value spots; and
- the lines are all drawn on the bed of the table and none of them cross?

[Monique]

R387 ...

147 apes: 74 orans (including Oliver) and 73 baboons with the number of baboons*number of apes = number balls initially led in one row/column by Oliver ... 73*147=10731

Suppose you number orans (except Oliver) and baboons in ascending order of size from 1 to k (k orans and k baboons)

then baboon i has 2*k*(10731+i) + k*k more balls than oran i. Summing this for i =1 to k you get
k*k*(21463+2*k) and that must be equal to the number of balls used by Oliver 10731*10731

therefore k must be a factor of 10731=3*7*7*73=147*73 ...

looking for a suitable k factor and knowing D_G and noticing that choosing 73 would yield 147 apes and a special relationship with 10731 ... I just checked the condition was met

[snookersfun]

right, my numbers are the same as Mon's

for the proof, I went for:
k= number of orangs (excluding Oliver) or babbons
n= 10731 Oliver's square

and here a picture first
mad1.GIF
a) # of balls for the orangs: basically k+1 inner squares of n^2 (green), then can add those lines on the side (dark blue) (need to sum 2kn, so basically k(k+1)n) and then add the sum of little squares 1 until k (red)
b) # of balls for the baboons: similar, but now k innner squares of (n+k)^2 instead of n^2 as above and sum light blue lines k(k+1)(n+k)

as a) = b) I now have the following equation (sum short for the 'little sum', as will cancel out anyway):

(k+1)n^2 + k(k+1)n + sum = k(n+k)^2 + k(k+1)(n+k) + sum

or in the end n^2 = k^2(2k+1+2n)

which after substituting n= 10731 leads to k=73.

[moglet]

R387

I got the same numbers as the others, but used the n(n+1)(2n+1)/6 formula and worked out the totals for 50 orangs and 49 baboons then for 100 orangs and 99 baboons, then looked at interpolating between the two. Like Monique, I went straight to 74/73 as an educated guess. This didn't take long, the extra time - many hours, was taken up trying to find a simple and direct formula to give the answer, without much success. Snookersfun's analysis is brilliant.