Puzzles with numbers and things

[Monique]

R401 ... Barry promotes his shop

On Monday ...
Barry the Baboon has got a great idea to promote his shop... he's organising a challenge for the visitors - they dont have to buy anything.
The idea is simple: he has arraged 169 snooker balls into a 13x13 square and each visitor who succeeds in rearranging those balls into two separate and different squares gets a free banana.gif

On Wednesday ...
Far too many visitors succeeded in rearranging this 13x13 square! Barry has spent a fortune in banana.gif and hasn't sold much Something needs to be done! So he decides to go for a bigger square, 15x15, 225 balls, and to ask the visistors to rearrange them into 3 different squares ... This time they will get 3 free banana.gif

On Friday ...
That's still not good! Oliver just left the shop having succeeded to rearrange the 225 balls into 3 squares -2x2, 5x5 and 14x14 respectively - and munching at his banana.gif. Charlie is now eager to play aswell and no doubt he will make it!
"Look" says Barry "I'll give you all the banana.gif I still have in stock for today, but please help me! I want to make this challenge much more difficult and have the visitors to try to rearrange some big square into 147 diffrerent smaller ones. Is that possible? How many balls do I need?"
"Ha!" answers Charlie "Yes it is possible. In fact it is possible for any number of squares you want ... and there are many, many solutions for each number"
Charlie smiles mischieviously and asks "How many banana.gif will you give me if I help you?"
"Eehrr, 10" says Barry.
"Pah!" exclaims Charlie "I'm not interested! Ask those TSFers "
The shop's door slams shut, and he's gone ...

Who can help Barry?

[Monique]

Now Snookersfun did it also. Congrats!

[moglet]

Perfect answer already by Moglet. Well done!

Well, yes, but after an initial blunder, thinking inside the snooker box!

[Monique]

Perfect answer already by Moglet. Well done!

[Monique]

R400

Rollie and Gwenny are playing snooker for fun, only they play an "ape" version of the game, meaning that they have a much bigger triangle of reds than usual ...

When they started their session none of them was on song and during the two first frames the highest break was ... 1.
Now Gwenny remarks: "That's funny from frame three on the highest break of the frame has each time been the sum of the highest breaks of the two previous frames ..."
"Eh yeah?" says Rollie, who's maths skills don't exactly match the ape's
"Yes", says Gwenny, "and if that pattern continues in the frame after the next one, one of us will eventually make a maxi"
"Oh yeah?" asks Rollie now slightly excited.
"And that means that this maxi high break will be higher than the sum of all high breaks we have made until now" peeps an enthousiastic Gwenny
"How do you know? wow! do you compute so quickly?" asks a rather nonplussed and admirative Rollie
"No need to compute" ... is Gwenny's answer

Now Rollie is truly

1. How many frames have they played?
2. How many reds do they have on the table?
3. Why is there no need for Gwenny to compute anything?

Note: the number of frames and the number of reds are the smallest possible fot this scenario to happen.

[snookersfun]

Battlechalks

they somehow go well with that new avatarVery nice, both!!

[moglet]

R399

Battlechalks

[snookersfun]

apparently I had a small glitch here yesterday...

I was meant to close my last open R.399, as Mon also sent a perfect solution.

So, the 'battlechalks' can come up on the thread now

Congrats to all!

[davis_greatest]

I was lucky enough to have an epiphanic moment with this one that meant I didn't have to do any lengthy trials and errors.

When Gwennie made her observation: 6 greens, 5 blues and 11 pinks (5+6=11)
For the previous (-1) turn: 5 greens, 4 blues, and 10 pinks (4+1=5)
For the last (+6) turn: 12, greens, 11 blues and 17 pinks (3+8=11)

Not sure if you did it this way, moglet, but the easiest way is to use the fact that any positive integer, and the sum of the digits of that number, both have the same remainder after dividing them by 9.

So if we write out equations that look only at the remainders after dividing by 9 ("modulo 9"), the three numbers and their respective sums of digits are equivalent, and you immediately get the answer you have put above.

[moglet]

R398

I was lucky enough to have an epiphanic moment with this one that meant I didn't have to do any lengthy trials and errors.

When Gwennie made her observation: 6 greens, 5 blues and 11 pinks (5+6=11)
For the previous (-1) turn: 5 greens, 4 blues, and 10 pinks (4+1=5)
For the last (+6) turn: 12, greens, 11 blues and 17 pinks (3+8=11)

[snookersfun]

updates:

R. 398: I have another correct answer by abextra, and Mon is kind of very much on her way to it Well done!
(and yes, solved perfectly now, so if anybody could put up the answer and explanation, we can close that round.)

R.399: moglet had no problem with the correct puzzle and I received a very nice picture full of chalks. Congratulations
edit2: another nice battlechalk picture, this time from abextra. Very well done

here is a hint in hidden text how to start out in that puzzle:
look at the upmost square with the 5 inside and figure its 'diagonal' options to position those 5 chalks

[snookersfun]

another gaaaah moment on here now. Unfortunately that attachment still appeared nicely, but was majorly wrong. Sooooorrrryyyy! Apologies for time wasted. Just replaced the faulty picture of R.399 with what it should be. Moglet made me aware of it, couldn't quite solve it that way

[davis_greatest]

R. 397: is now perfectly solved by all the regular pundits, congratulations d_g, moglet, abextra and Mon.
Would somebody please put up his/her solution with a short explanation.

Let price per set in £ = 10p+q (0<=q<=9), so total money = (10p+q)² = 20p(5p+q) + q²
q² must have first digit odd, giving only q² = 16 or 36, so towel costs £6 and Oliver must give something worth half of the extra £4 in value he has, so clamp is worth £2

[snookersfun]

R. 397: is now perfectly solved by all the regular pundits, congratulations d_g, moglet, abextra and Mon.
Would somebody please put up his/her solution with a short explanation.

R. 398 update: the boys are in first on that one, d_g two ways, a cheating vb code for excel and the way he always asks us for, and moglet reached the solution as well. Very well done!

and maybe one more last one for now:

R. 399: Placing Triangle chalks, or battleships

In the following 10x10 grid please position pieces of triangle chalks in straight lines of 4 (1 each), 3 (2 each), 2 (3 each) and 1 (4 each) in such a way that none of the conformations touch neighbouring ones (not even diagonally).
chalk.bmp
The numbers given in the grid show the pieces of triangle chalk which can be found a) as the sum of those in each respective row and column and also b) as the sum of those in each of the two lines diagonally crossing through the respective numbered spot. No chalks can be placed on the numbered spots.

answers by PM please

[snookersfun]

thank you moglet, very nice picture

and update to R. 397: this was first solved perfectly by d_g (thought I just mention it, as the previous rounds seem to take him forever), followed by moglet, and abextra and Mon need to make some small adjustments but are basically there as well.

So with that we can tighten that clamp a bit and:

R. 398: filling boxes

Coming back home the apes meet up with Gordon and Gwenny and look over their boxes of spare balls. Balls are generally boxed by colours and each box can contain up to 30 balls.
Gordon, Charlie and Oliver each take an open box already containing some green, blue and pink balls respectively (all different numbers) and then take turns adding one ball each of their particular colour fished out from the 'messy box' (a huge box where leftover balls are often initially thrown in regardless of colour for later sorting) into their boxes.

Gwenny, running around between them and peeping into all the boxes suddenly pipes up:
Oh, this is interesting; on this round of turns the amount of pink balls Oliver has accumulated in his box sums up to the sum of the digits of the amount of blue added to the sum of the digits of the amount of green balls Charlie and Gordon have amassed.
Furthermore, on the round before one could have observed the same for the amount of Gordon's balls (then being the sum of the digits of the two other ape's balls) and in 6 more rounds, finally Charlie's sum will equal the sum of the digits of the other two balls.

So, how many balls did each ape have when Gwenny made the observation?

answers and explanations by PM please