Puzzles with numbers and things

[snookersfun]

breaking news

moglet just filled in all scores correctly. Well done!

[snookersfun]

Mon and abextra solved R.409 as well... Congratulations to all!

here is another one like it:
R. 410 keeping scores-2

Can you fill in this round's sheet, if Charlie gives you the following information?

1) No number appears more than once
2) Row 1) the tens digits are all different
3) Row 2) numbers are even
4) Row 4) the sum of the 5 numbers is 178
5) Row 5) numbers are prime
6) Column S) the sum of the 5 numbers is 60
7) Column C) two of the numbers are bigger than 19
8) Column E) numbers are increasing (equally spaced) from top to bottom
9) The product of S3 and E3 is between 600 and 700
10) The sum of C1 and R5 equals the sum of R1 and C5
11) O1 is midway between C4 and R4
12) O5 is midway between C2 and R2
13) C3 is midway between S1 and O1, and R3 is midway between O1 and E1
14) Each column and row contains one (and only one) multiple of 11
15) There is one triangular number in every column

Answers by PM please

[Monique]

Thanks Snookersfun.

I'd like to have a few clarifications...

1. May we assume that O3 is empty like in the former round or not?
2. In condition 7, "two" means exactly two or at least two? I suppose the first.
3. In condition 2, can we assume there actually is a tens digit for all numbers in that row?

[snookersfun]

Thanks Snookersfun.

I'd like to have a few clarifications...

1. May we assume that O3 is empty like in the former round or not?
2. In condition 7, "two" means exactly two or at least two? I suppose the first.
3. In condition 2, can we assume there actually is a tens digit for all numbers in that row?

lol, and the last round went so well, without clarifications and blunders.

1) yes, sorry, O3 is empty again, use the same template
2) seemed to turn out to mean exactly two
3) not sure about that, better not to assume, the solution will be unique in any case (can't remember this clue helping me much)

Hope that helped

[snookersfun]

update: R. 410 is solved perfectly by moglet and abextra so far. Mon should get an honourable mention for finding all my blunders for now ...and meanwhile also earned her banana fraction, another perfect solution in. Congrats to all!

Solutions to R.409 and 410 can come up on the thread now.

with that on to...

R.411 summing balls

Place the snooker balls (ball values from 1 to 7) into the following grid, so that each row and column contains at most one of each kind of balls (therefore leaving some places blank).

sumballs.bmp
The numbers at the side of the diagram indicate the sums of ball values in all blocks of consecutively placed (touching) balls in a row or column and the order of those blocks. Single ball values are listed as well.

Answers by PM please

[Monique]

R412 ... Betting freaks

The apes and Rollie had a very frustrating day of snooker, nothing worked for them! It ended up in disaster when Charlie knocked the blue off the table so violently that he broke a trophy that was on display in the club! After which the referee called it a day and went taking the blue and all the reds with him!


So they are now trying to get some steam out in order to go from to again. For that purpose, each of them has chosen one ball - black for Charlie, brown for Oliver, green for Gordon, pink for Gwenny and yellow for Rollie - and they just knock them around the table randomly as hard as they can. When all of them have "played" they look at the pattern on the table and start over again :snooker: ... silly but efficient

Barry is watching with his mate Bongo (another baboon, not "our" Bongo) and they are taking bets. Each time the blue spot will be in the triangle formed by the black, brown and green Barry will give Bongo a banana while each time the blue spot will be in the rectangle formed by tracing imaginary parallels to the cushions from the pink and yellow Bongo will give Barry a banana.

If that game goes on all night - that means many, many rounds of "steaming out" - who do you expect from Bongo or Barry to have "earned" more bananas and why?

[abextra]

update: ...

Solutions to R.409 and 410 can come up on the thread now.

Round 409 :snooker::

11 . 17 . 41 . 59 . 61
12 . 18 . 36 . 57 . 63
13 . 19 . ... . 54 . 66
14 . 20 . 33 . 52 . 70
15 . 21 . 32 . 46 . 75

[moglet]

R410 solution

[snookersfun]

Thank you for all the solutions up on the thread!

The solution to R.411 can come up as well now, as it is perfectly solved by moglet, abextra and Mon. Congratulations to all of you!

So now to a new one. Bear with me, as I am nearly sure to make this unintelligible, miss clues or add some blunders in. Therefore feel free to ask questions on the thread, by PM or however you see fit to do!

R. 413 Balls in and out

Charlie built a new smart toy for the apes, a wooden construction, a thin box of about 1 ball width, represented 2-dimensionally in the drawing. Balls go in at the top through 16 holes lined up close to each other, travel down vertically along the vertical grid lines (if not deflected) and come out at the bottom.
molehills-1.bmp

The highlights of the whole construction are the deflectors (examples shown at left hand top), bridge like structures, which are placed in each row (amount, order and length shown at the left side of the scheme- the deflectors may touch but not overlap for a row). Balls landing on the sloped parts of it are deflected downside into the path closest to the respective endpoint of the deflector. Deflectors of even lengths have half the balls going to either side, if balls are landing on its midpoint (therefore they can be only placed midpoint under an even number of balls (possibly combined values).

The numbers of balls in (through each specific hole) are shown on top and the amounts of balls out are shown numerically at the bottom of the scheme.

Help Charlie place the deflectors in order to have the balls roll out at the bottom as shown.

Answers by PM please.

[abextra]

Thank you for all the solutions up on the thread!

The solution to R.411 can come up as well now, as it is perfectly solved by moglet, abextra and Mon. Congratulations to all of you!

Round 411 solution :snooker:

[snookersfun]

R.413 late update: solved by Mon, abextra and moglet. Very well done!

so this can come up as well now

[moglet]

R413

So preoccupied with R412 I nearly missed it!



Re R412, this is a fascinating puzzle/proposition. The answer is not as obvious as it may seem. Lots of geometry, coordinates and how you interpret them. No posts on the forum about it yet, but, a big, big thanks to Monique for opening "the box".

[moglet]

Re R412, this is a fascinating puzzle/proposition. The answer is not as obvious as it may seem. Lots of geometry, coordinates and how you interpret them. No posts on the forum about it yet, but, a big, big thanks to Monique for opening "the box".

[abextra]

Re R412, this is a fascinating puzzle/proposition. The answer is not as obvious as it may seem. Lots of geometry, coordinates and how you interpret them. No posts on the forum about it yet, but, a big, big thanks to Monique for opening "the box".

Hi, Moglet! Did you solve this question?

I found the answer - and I have to say it surprised me - but my reasoning was wrong, so I'm afraid it was just a fluke... probability is not for me.

[moglet]

Hello back abextra, I'm still working on it but the proofs and analysis are pretty well there. Unfortunately Monique has dropped off the grid for the moment and so far is not replying to "email addresses" or PM's while she is away. If I do hit guaranteed gold with my stuff I'll post directly to the thread.

I'll send you some stuff as soon as I can.