Puzzles with numbers and things

[chasmmi]

Round 89

A point to anybody who PMs me the following by midnight Tommorow Korean Time (I think 3pm on Saturday for you all)

1) I would like 3 perfect numbers

2) I would like 2 sublime numbers

3) I would like 3 weird numbers

I may award half points

[davis_greatest]

What a perfectly sublimely weird question! But slow down, everyone, with the question-asking ... that's three open rounds at the moment I'll wait for one if not two of them to close, before I continue asking questions.

Round 86 closes tonight! And snookersfun's round 88 is open until Sunday, I believe.

[davis_greatest]

Round 89

A point to anybody who PMs me the following by midnight Tommorow Korean Time (I think 3pm on Saturday for you all)

1) I would like 3 perfect numbers

2) I would like 2 sublime numbers

3) I would like 3 weird numbers

I may award half points

If you find me 3 sublime numbers, chasmmi, I'll give you 5 points myself!

[Snooker Rocks!]

Without taking what DGE just said in mind, here is ROUND 90.

The festive season is approaching, and we've all heard the following poem/song/thingy:

'On the first day of Christmas,
my true love sent to me
A partridge in a pear tree.

On the second day of Christmas,
my true love sent to me
Two turtle doves,
And a partridge in a pear tree.

On the third day of Christmas,
my true love sent to me
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the fourth day of Christmas,
my true love sent to me
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the fifth day of Christmas,
my true love sent to me
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the sixth day of Christmas,
my true love sent to me
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the seventh day of Christmas,
my true love sent to me
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the eighth day of Christmas,
my true love sent to me
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the ninth day of Christmas,
my true love sent to me
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the tenth day of Christmas,
my true love sent to me
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the eleventh day of Christmas,
my true love sent to me
Eleven pipers piping,
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the twelfth day of Christmas,
my true love sent to me
Twelve drummers drumming,
Eleven pipers piping,
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree!'

My question is, how manny presents did she get?

I think 1/2 marks are fair for anyone who goes and does the addition, but full marks will be awarded to anyone who uses another, neater method.

Answers by PM please, deadline is 20:00 on Sat. 23rd December.

[April madness]

Round 89

A point to anybody who PMs me the following by midnight Tommorow Korean Time (I think 3pm on Saturday for you all)

1) I would like 3 perfect numbers

2) I would like 2 sublime numbers

3) I would like 3 weird numbers

I may award half points

I shall go back to school and shoot my maths teacher. No idea what a perfect or sublime numbers is!

[davis_greatest]

I shall go back to school and shoot my maths teacher. No idea what a perfect or sublime numbers is!

Don't worry, April - these aren't usually mentioned at school Probably they'll be in Wikipedia... if not, a google search

[April madness]

Don't worry, April - these aren't usually mentioned at school Probably they'll be in Wikipedia... if not, a google search

Not that I worry or I will answer to the question

[davis_greatest]

Reminder of open rounds

As there are now 4 rounds open, here is a reminder of the open ones. No other question please, while these are open! I will resume, posting from round 91 tomorrow...

Round 86

Oliver, my pet orang-utan, is playing snooker with Angles McBum, on a perfectly rectangular snooker table with a playing area exactly 6 feet wide by 12 feet long. (The Statman, I know that this is slightly larger than the official playing area allowed, but that is the size of this snooker table and there is nothing I can do about it. Throughout this question, also, ignore the width of the balls - i.e. assume that the balls have zero diameter.)

Oliver and Angles are down to the final pink and black and Oliver has trapped Angles in a snooker, with the cue ball at least 5 feet away from the object ball (the pink).

The cue ball happens to be touching a cushion (not the top cushion) and is lying a whole number of feet (at least one foot) from the nearest corner pocket (or pockets).

The pink is not touching a cushion, but happens to lie a whole number of feet (at least one foot) from each cushion. This means that if you measured from the pink to the nearest point of each of the 4 cushions in turn, each time it would measure a whole number of feet exactly. (We are treating the table as having 4 cushions, one at each end and one at each side - ignore the fact that the middle pockets break up the cushions on each side.)

Angles is known for his superb knowledge of the angles on a snooker table and is extremely good at getting out of snookers. Furthermore, he is adept at using sidespin to such an extent that he can make the cue ball rebound off a cushion at whatever angle he chooses - the cue ball does not need to rebound at the same angle as it hit the cushion. (He can do this even when playing from tight on the cushion rail.)

Angles decides to play a one cushion escape - not off the top cushion though (and not off the cushion that the cue ball was already touching, as this would be impossible). Once he has chosen the cushion off which to rebound, he uses sidespin if necessary in such a way as to make the cue ball travel the shortest total distance possible, using that cushion, before the cue ball hits the pink. (This distance includes the travel to the cushion plus the travel from the cushion until striking the pink - i.e. add the two together.)

Note: this does not necessarily mean that the escape that Angles chooses makes the white travel the shortest possible distance of all possible escapes - there might be a shorter escape off another cushion. What it means is that, once the cushion for the escape has been chosen, Angles chooses the point of contact on that cushion in such a way as to make the cue ball travel the shortest distance.

Angles plays the shot perfectly, as expected, and strikes the pink full ball. It also just happens that the total distance that the cue ball travels in this escape, until striking the pink, is again exactly a whole number of feet (but not a multiple of 5 feet, so not 5, 10, 15 etc).

If Angles had chosen a different route and instead played the one cushion escape off the top cushion with no sidespin (so the cue ball rebounds at the same angle as it strikes the cushion), how far would the cue ball have travelled from being struck until hitting the pink?

Answers by Private Message please. Deadline is 23:00 GMT tonight, Friday 21 December.

There was also a clue to round 86, here: http://www.thesnookerforum.com/showp...ount_1107.html


snookersfun's round 88

I’ll put up another question, round 88:
You should try to arrange the numbers 1,2,3,….n (two of each) in such a way, that there is one other number between two ones, two other numbers between two 2s, three other numbers between two threes, etc.

Example: for n=4 i.e. using 1,2,3,4 twice each: 4,1,3,1,2,4,3,2
Try to get as high a sequence as possible in this way. Bids for number of n reached (so ½ the digits in the sequence) can be put up on the thread until Sunday 24.12. 12:00pm GMT. After that I’ll ask for the actual sequences.

chasmmi's round 89

Round 89

A point to anybody who PMs me the following by midnight Tommorow Korean Time (I think 3pm on Saturday for you all)

1) I would like 3 perfect numbers

2) I would like 2 sublime numbers

3) I would like 3 weird numbers

I may award half points

Snooker Rocks!'s round 90

Without taking what DGE just said in mind, here is ROUND 90.

The festive season is approaching, and we've all heard the following poem/song/thingy:

'On the first day of Christmas,
my true love sent to me
A partridge in a pear tree.

On the second day of Christmas,
my true love sent to me
Two turtle doves,
And a partridge in a pear tree.

On the third day of Christmas,
my true love sent to me
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the fourth day of Christmas,
my true love sent to me
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the fifth day of Christmas,
my true love sent to me
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the sixth day of Christmas,
my true love sent to me
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the seventh day of Christmas,
my true love sent to me
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the eighth day of Christmas,
my true love sent to me
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the ninth day of Christmas,
my true love sent to me
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the tenth day of Christmas,
my true love sent to me
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the eleventh day of Christmas,
my true love sent to me
Eleven pipers piping,
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree.

On the twelfth day of Christmas,
my true love sent to me
Twelve drummers drumming,
Eleven pipers piping,
Ten lords a-leaping,
Nine ladies dancing,
Eight maids a-milking,
Seven swans a-swimming,
Six geese a-laying,
Five golden rings,
Four calling birds,
Three French hens,
Two turtle doves,
And a partridge in a pear tree!'

My question is, how manny presents did she get?

I think 1/2 marks are fair for anyone who goes and does the addition, but full marks will be awarded to anyone who uses another, neater method.

Answers by PM please, deadline is 20:00 on Sat. 23rd December.

[davis_greatest]

Round 86 "They call him Angles" completed!

Congratulations to snookersfun and abextra for correctly finding that Angles McBum would have made the cue ball travel 13 feet, had he played off the top cushion. This, in fact, is the same distance as the escape he chose, which was off the bottom cushion!

The cue ball must be on the side cushion - but it is impossible to determine where - and the pink ball must be 1 foot from the opposite side cushion. Although it is impossible to determine exactly where the white and pink are, we don't need to know in order to solve the question - the answer to the question MUST still be 13 feet.

In fact, the white could be anywhere from 1 to 11 feet from the top cushion - and the pink must be an equal distance but from the bottom cushion!

The key to solving this is to imagine that the cue ball travels x feet up/down the table (possibly off a cushion and back again) and y feet sideways (possibly off a cushion and back again). That is, we are separating the motion of the cue ball into its up/down movement parallel to the side cushions, and its left/right movement parallel to the top and bottom cushions.

Let the total distance that the white travels for McBum's original escape be z feet.

I told you that x, y and z are all whole numbers.

Then Pythagoras's theorem for right-angled triangles tell us that

x^2 + y^2 = z^2

So we need to find Pythagorean triples x,y,z that satisfy this equation. The smallest ones are

3,4,5 (3^2 + 4^2 = 5^2) and multiples of that - which we are told in the question are not the solution

AND

5,12,13 (5^2 + 12^2 = 13^2) - which is therefore the only possibility.

All other Pythagorean triples are too big to fit on a snooker table without requiring more than one bounce off a cushion.

The cue ball cannot travel 12 feet sideways on a snooker table without touching 2 cushions, so must go 12 feet down and up; and 5 feet sideways.

This means that McBum plays his shot off the bottom cushion.

Since the route off the bottom cushion down and then up is 12 feet, and 5 feet sideways, then if McBum were instead to play off the top cushion, the route up and then down must also be 12 feet (because the table is 12 feet long!) and 5 feet sideways.

So, using Pythagoras and the fact that 5^2 + 12^2 = 13^2, the route off the top cushion must again be 13 feet!

A deceptively simple question! And one that works beautifully because a snooker table is 12 feet long.


SO HERE IS THE SCOREBOARD AFTER ROUND 87 BUT BEFORE ROUNDS 88, 89 AND 90

snookersfun………....………….…..44½
abextra...............................26
davis_greatest.....................19
Vidas..................................12½
chasmmi..............................10
elvaago................................9
Sarmu..................................8
robert602.............................6
The Statman…………………. …...…5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
April Madness........................1
Ginger_Freak.........................1
Snooker Rocks! .....................1

[chasmmi]

My round is now over with point for Snooker rocks, Abextra, Davis Greatest and Elvago.

[davis_greatest]

SO HERE IS THE SCOREBOARD AFTER ROUND 90 BUT BEFORE ROUNDS 88 AND 90

snookersfun………....………….…..44½
abextra...............................27
davis_greatest.....................20
Vidas..................................12½
chasmmi..............................10
elvaago...............................10
Sarmu..................................8
robert602.............................6
The Statman…………………. …...…5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
Snooker Rocks! .....................2
April Madness........................1
Ginger_Freak.........................1

[davis_greatest]

Round 91 - 1000 bananas on Red or Black

As a reminder, any bids to snookersfun's round 88 above must be made by noon GMT today (24 December). Since no one (other than davis_greatest) has bid and since the 2nd highest bidder will also score, there could be an easy point for someone!

The deadline to Snooker Rocks!'s round 90 has now expired - no doubt Snooker Rocks! will soon announce the result.

So, let's proceed to....

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.

.

[snookersfun]

I am back now, missed some easy rounds it seems.
Just to remind you, that my deadline will expire soonish (5 1/2 h to go).
Nobody can do a n=7? (I don't expect anybody to beat the 80, but 7 should be doable)

[davis_greatest]

Even the answer is odd!

By the way, it's impossible to find any solutions to snookersfun's problem if n is 1 or 2 more than a multiple of 4 (i.e. of the form 4k+1 or 4k+2). That's why 1, 2, 5, 6, 9, 10, 13, 14, 17, 18 etc are all impossible.

The reason is that, for any valid sequence, any pair of odd numbers must either both appear in even positions or both appear in odd positions in the sequence* (since they have an odd number of numbers in between them!). Since the total length of the sequence is even (as it consists of n pairs), the sequence must have the same number of odd positions as even positions. Therefore, the number of odd pairs must be even. So n=4k+1 or 4k+2 can't work because such sequences would have an odd number of odd pairs. How odd!


* For example, in the 4,1,3,1,2,4,3,2 sequence, the odd pair 1,1 appears in the 2nd and 4th positions (both even) and the odd pair 3,3 appears in the 3rd and 7th positions (both odd). Note that each pair is the same - i.e. both numbers in the pair are in even positions, or both numbers in the pair are in odd positions.

[Snooker Rocks!]

Ok round 90 has finished.

Full points to: davis_greatest, Robert602 and abextra.

Half points to: Ginger Freak