Puzzles with numbers and things

[April madness]

Hehe Mike, if I wouldn't know it's Christmas Eve today, I'd think it's mid-April and world snooker championship starts tomorrow! The grass is green, no snow, was sunny today and +6 C! But not as warm as in August

[davis_greatest]

Davis greatest ever type character,

What are you getting from the bearded guy this year? The snooker fan's guide to calculus perhaps?

Actually, I'm hoping that the bearded guy will give me his new anthology of his best quotations. The book, entitled "Pot as many balls as you can" apparently begins with the classic "Where's the white ball going?" and finishes with the legendary "It's there... it's THERE!". At the back of the book is a bonus section showing shots to play in commonly occurring positions of the table in tournament play - for instance, when you find 3 cues lined up on the table in a straight line to the corner pocket, with 19 touching reds in an S-shape and a wicker basket balanced on a block of chalk over the middle pocket. What are you hoping to find in your stocking in the morning?

[davis_greatest]

oh, and here is Abextra's number:
7 3 6 2 5 3 2 4 7 6 5 1 4 1

and, I am adding a solution for 8 from Chasmmi, who is in Seoul and doesn't know, what is the time here... so, in the Christmas spirit, have a point as well very nice!

7 8 2 3 6 2 5 3 7 4 8 6 5 1 4 1


and here is a solution for 20 numbers done with d_g's program:
20,18,19,15,11,17,10,16,9,5,14,1,13,1,12,5,11,10,9 ,15,18,20,19,17,16,14,13,12,8,4,7,3,6,2,4,3,2,8,7, 6

he might want to put up his highest number solution later on...

OK, here's one with 84:

84, 82, 83, 79, 77, 81, 74, 80, 71, 69, 78, 66, 64, 76, 61, 75, 58, 56, 73, 53, 72, 50, 48, 70, 45, 43, 68, 39, 67, 36, 33, 65, 30, 27, 63, 23, 62, 17, 14, 60, 15, 59, 7, 4, 57, 5, 11, 55, 4, 54, 7, 5, 52, 14, 51, 17, 15, 49, 11, 23, 47, 27, 46, 30, 33, 44, 36, 39, 42, 43, 45, 48, 50, 53, 56, 58, 61, 64, 66, 69, 71, 74, 77, 79, 82, 84, 83, 81, 80, 78, 76, 75, 73, 72, 70, 68, 67, 65, 63, 62, 60, 59, 57, 55, 54, 52, 51, 49, 47, 46, 44, 42, 41, 38, 40, 37, 34, 32, 29, 26, 35, 25, 16, 13, 24, 31, 8, 6, 3, 28, 12, 9, 3, 10, 6, 8, 22, 13, 21, 16, 20, 9, 19, 12, 10, 18, 26, 25, 29, 24, 32, 34, 38, 37, 41, 40, 35, 31, 28, 22, 21, 20, 19, 2, 18, 1, 2, 1.

Here's the code in Visual Basic, which pastes the solution into Excel if you name a cell "number_of_pairs" on an Excel worksheet and type a number into it (in the above case, I typed 84). The calculations are done in the 3rd of the 4 blocks of code. The other blocks just define variables, output the numbers etc.

Option Explicit
Public number_of_pairs As Integer
Public digits() As Integer
Public location() As Integer
Public outputcount As Integer

Sub search()
number_of_pairs = Range("number_of_pairs")
ReDim digits(2 * number_of_pairs) As Integer
ReDim location(number_of_pairs) As Integer
Range("A3", Selection.End(xlDown)).ClearContents
set_number (number_of_pairs)
End Sub

Sub set_number(k As Integer)
For location(k) = 1 To (2 * number_of_pairs - k - 1)
If (digits(location(k)) = 0) And (digits(location(k) + k + 1) = 0) Then
digits(location(k)) = k
digits(location(k) + k + 1) = k
If k = 1 Then
output
End
Else
set_number (k - 1)
End If
digits(location(k)) = 0
digits(location(k) + k + 1) = 0
End If
Next
End Sub

Sub output()
For outputcount = 1 To 2 * number_of_pairs
Range("number_of_pairs").Offset(outputcount + 1, -1) = digits(outputcount)
Next
End Sub

[snookersfun]

btw. how long did it tak your PC to come up with the answer for 84? While 40 comes up in a sec, 60 never seems to finish for me...

[davis_greatest]

btw. how long did it tak your PC to come up with the answer for 84? While 40 comes up in a sec, 60 never seems to finish for me...

It's funny, because sometimes one number might take a long time, but trying a larger number can be quicker.

40 appears for me virtually as I press the button, 67 takes about 5 seconds on my computer, 76 takes about 15 seconds, and 84 takes about a minute.

But some numbers less than 84 can take much longer - so long I haven't waited! Best thing to do is wait a few seconds, and if nothing comes up very quickly, stop it and try a different number.

[davis_greatest]

Merry Boxing Day to all

Two correct answers have arrived by Private Message to Round 91. Any further answers should be posted here on this thread.

The round will close once the next correct answer is posted.

Round 91 - 1000 bananas on Red or Black

Gordon and Oliver like to gamble at cards, and they play a little game. Gordon starts with 1000 bananas to gamble. Oliver, the banker (and a good banana tree-climber), has a virtually unlimited supply of bananas with which he can accept bets.

Charlie, the dealer (who is not betting) takes an ordinary pack of 52 playing cards and gives them a good shuffle. He then turns over each card, one at a time, until he has gone through the whole pack.

Before each card is turned over, Gordon and Oliver bet on its colour. Gordon will always bet Red, and Oliver will always bet Black. Each time, they will bet an amount equal to one-tenth of the bananas that Gordon owns before the card is turned. So, if the card turned is black, Oliver wins one-tenth of Gordon's bananas. If the card is red, Gordon wins that number of bananas from Oliver.

Therefore, the bet on the first card is for 100 bananas because Gordon starts with 1000 bananas. Incidentally, all my apes are very good at fractions and so betting fractions of a banana, if necessary, is no problem.


Now, here's where you come in. Before they start the game, you are allowed to back one of the two apes (Gordon or Oliver), and bet on which ape you think will have more bananas than he started with, once all 52 cards have been turned.

If you want to participate, you must gamble 1000 bananas of your own - and here is the deal:

- if, at the end, the ape you back has more bananas than he started with, you get all the "bananas profit" that he won (i.e. you win the bananas that he had won from the other ape during the game);

- but if, at the end, the ape you back has fewer bananas than he started with (or the same number), you lose your whole 1000 bananas-stake.


Of course, you want to maximise your bananas.

So....

(a) Should you participate and back an ape?

(b) If the answer to (a) is "yes", then which ape do you back? How many bananas would you expect to win if you back that ape?

(c) If the answer to (a) is "no", then why shouldn't you back an ape? How many bananas would you expect to lose if you backed an ape?


Initially, at least, answers to this question - with explanation - may either be:
- posted on this thread (by those outside the top 4 on the scoreboard); or
- sent to me by Private Message (by those inside the top 4).
I reserve the right to paste here on the thread any answers sent to me by Private Message.

.

[Snooker Rocks!]

My round is over. Point are to awarded to:

chasmmi
snookerfun
elvaago
davis_greatest
abextra

(listed in the order I received answers)

[davis_greatest]

Perhaps a little meanly, I'll award half a point each to the solvers of round 92. We don't want to give these valued points away too easily, or everyone will be getting a banana
davis ebeneezer scrooge greatest

SO HERE IS THE SCOREBOARD AFTER ROUND 92 BUT BEFORE ROUND 91

snookersfun.........................45
abextra...............................29½
davis_greatest.....................23½
Vidas..................................12½
chasmmi..............................11½
elvaago...............................10½
Sarmu..................................8
robert602.............................7
The Statman.........................5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
Snooker Rocks! .....................2
Ginger_Freak.........................1½
April Madness........................1

[abextra]

Round 91

Is it possible to win a fraction less than 230 bananas by backing Oliver?

Every time the card turned is red, Gordon recieves 10% of the bananas he has before turning, it means the amount of his bananas will be multiplied by 1,1. If the card turned is black, he loses 10% of his bananas,it means the amount of his bananas will be multiplied by 0,9. As there are 26 red cards and 26 black cards in the pack, the total number of Gordon's bananas after turning over all 52 cards will be

1000 * 1,1^26 * 0,9^26 = ~770,04 bananas.

So Gordon ends with ~770 bananas and Oliver will have ~230 bananas more than he started with.

[davis_greatest]

That's a perfect answer! ... meaning that round 91 is closed, with points awarded to chasmmi and snookersfun (both for submissions by Private Message) and abextra for the answer above.

SO HERE IS THE SCOREBOARD AFTER ROUND 92

snookersfun.........................46
abextra...............................30½
davis_greatest.....................23½
Vidas..................................12½
chasmmi..............................12½
elvaago...............................10½
Sarmu..................................8
robert602.............................7
The Statman.........................5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
Snooker Rocks! .....................2
Ginger_Freak.........................1½
April Madness........................1

ROUND 93 ...

... should hopefully follow before I go out in the next hour

[davis_greatest]

Round 93 - The Spirit of Christmas Present

At my post-Christmas party, abextra, April Madness, austrian_girl, Charlie, chasmmi, elvaago, Ginger_Freak, Gordon, Oliver, robert602, Sarmu, Semih_Sayginer, Snooker Rocks!, snookersfun, The Statman and Vidas are all dancing and running around my circular dining table to the music, getting rather drunk on my Cointreau and other spirits, while playing musical chairs. What a funny sight it is!

When I turn off the music, they each rush to a seat and sit down (equally spaced around the table), thinking that there will be one seat too few, while I stand up watching. In fact, there is just the right number of seats to seat them all, and I have laid out in advance a Christmas present at each place on the table, each labelled with one of their names.

Now, it turns out that in the clamour, no one at all is sitting at the right place for his or her present! I see that everyone is tired after all the dancing and running and so, to make things easier in my generous Christmas mood, I decide to rotate the whole table itself in such a way as to make as many presents as possible match up with their intended recipients!

Little do I know, however, that naughty Charlie knew what I had been planning, and he had secretly told everyone in advance where to sit so that no one would be sitting initially by his or her present! Not only that, but he had told everyone to sit in particular places such that, no matter how I subsequently rotated the table, the fewest possible people / apes could be matched up with their presents! Naughty, naughty chimpanzee!

So... after I rotated the table, how many presents did I manage to match to their intended recipients?

This time, answers - and explanations - may be posted either on this thread or sent to me by Private Message.

[davis_greatest]

Clarification - rotating the table is quite tiring and I can only rotate it once (but I can rotate it as far round as I want). No one is able to take a present until I have finished rotating the table!

[elvaago]

Charlie would probably be able to seat everyone in such a way that you could only match up one present with one person.

[Snooker Rocks!]

I've found a way with only 2 (I think ), that's just been PMed.

[davis_greatest]

Round 93... update and hints

Well, I've had a variety of answers sent to me - some better than others, but none wholly satisfactory yet

Some are saying that

(a) Charlie should be able to find a way to make people sit so that I can only line up one present with its intended recipient;

while some are saying that

(b) I should be able to match two presents to their intended recipients.

Well... I like the answers (b) more BUT the only answers I've had showing two recipients have given a single specific way of arranging the partygoers at the table - with it then possible to match 2 presents.

For those answering (b) (which is correct ), you need to show that Charlie could not possibly find any arrangement that would prevent me from matching more than one present. It is not good enough for you simply to state one particular arrangement that allows me to match 2 presents - that does not prove that there is no other arrangement for which I could not match 2 presents. (Otherwise, you might just as well seat everyone to the right of his/her present and say that the answer to the question is 16.)