Puzzles with numbers and things

[davis_greatest]

Here, when I have the time (and when I can find this thread - like many others, I am new here and still learning my way around), I shall post puzzles, problems etc, often with a mathematical slant. Others are welcome to do the same.

Here is a (relatively easy, I think) starter for ten.

Question 1

There are 147 of us in my local snooker league, which I run. During the season, we each play everyone else in the league once (in a best-of-one-frame match) and the winner of each match is allowed to take away a red ball at the end. (The club owner, who has many balls, does not seem to mind.)

At the end of the (very long) season, the top two players are invited to play in a Supersnooker final, on a huge table I had made especially. Instead of using the normal 15 reds, the reds that all the players collected over the season are put in a big triangle (so, if there were 15 reds, it would be just like snooker, but there are actually many more than that).

At the end of this year's final, I set up the table again and invited everyone in the league to take, in turn, a whole row of reds from the triangle to keep as a souvenir. Sadly, when it was my turn, there were no reds left.

How many of my fellow players suffered a similar fate and went home empty-handed?

[Robert602]

Just you

I actually worked this out the long way (0.5n(n+1), quadratic forumulae and all) before I realised it was just triangular numbers forwards and backwards.

There are 146+145+144+...+1 matches, and therefore the same number of balls. With that many balls you can of course make a triangle of 146 rows (1 ball, 2 balls... 146 balls). You must've been last in line DGE.

Do I get a bonus point for saying that the maximum possible break (free balls aside) in the final would've been 84707?

[davis_greatest]

Indeed, I was the only unfortunate one. Of course, it would have been the same answer regardless of how many players were in my league (provided there are at least two - with fewer than two it would not be very exciting and the final would be a morose affair).

I am afraid that I cannot award you a bonus point for saying that the maximum break (free balls aside) would have been 84,707. Had you said that the maximum break (free balls aside) would have been 85,875 (=147x146/2 x 8 + 27), maybe it would have been a different matter....

[Robert602]

D'oh!

[Mitsuko]

I did Maths A level last year and have no idea what you are on about

So I repeat Rob and say

[Robert602]

I did Maths A level last year and have no idea what you are on about

So I repeat Rob and say

As DGE reminded me earlier, I'm supposed to be doing a maths degree from september, and I seem to have trouble with numbers once I run out of fingers to count on. Very worrying .

It's a shame applied maths doesn't suit itself to interesting snooker puzzles, I'm alright at that .

I'll see if I can find a tricky one for you tomorrow DGE .

[The Statman]

According to my page, my average daily activity is 27½ posts.

I don't recall posting half a message!

[abextra]

I liked your question,DG. Will there be more?

[davis_greatest]

Certainly, abextra Thank you.

I am about to go and get ready to watch the football, so don't have time this very minute to think up a new question, but in the meantime I shall quickly repost here a few of the questions I posed in the BBC Five Live thread which I don't think were ever successfully answered (some of which I have changed slightly)!

And so...

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Q2) 20 years from now, you discover in your attic a 40 inch cue, which you were given as a boy by a famous player at the end of his match in the world snooker championships back in the year 2006. Unlike normal cues, it is of uniform thickness (and density) throughout its length and it weighs exactly 40 ounces. It also, bizarrely, has no tip, although it appears that it may once have had one (possibly, many, many tips) – which appears to have been bitten off.

You decide to chop it up because you need to be able to weigh any item that is a whole number of ounces (up to 40oz; i.e. 1oz, 2oz, 3oz.,…,40oz) using just balancing scales and pieces of wood from the cue.

i.e. you can put the item you want to weigh on one side of the scales and can put bits of the chopped up cue on either side as you choose.

You may only cut the cue into four pieces. How long should each piece be?


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Q3) This is my variation on a classic puzzle:

You are stuck in the centre of the Magic Forest and arrive at the Eightfold Paths with the Three Witches. You know that seven of the paths lead to a lifetime playing qualifiers and that the other leads to the 888.com World Title and a high-pitched voice (which you don't mind), but you do not know which path is which.

You also know that only one of the witches always tells the truth, and that the other two always lie, but again you do not know which witch is which. You may ask each witch one question and one only, to which the old hag may only reply Yay or Nay.

What do you ask each witch?


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Q4) This one WAS answered, by The Statman! (But he cheated and used Excel ) I offer it again here for anyone else who did not see the answer and would like to have a go. We need the answer for this before we move on to question 5!


On the planet of Crucibilis, 147 million and 991 light years from Earth, they like to play Nookers. It is played on a large green table (bigger than Earth), with one each of a white ball, yellow, green, brown, blue, pink and black balls, and a big pack of reds that at the start of a frame are arranged in a square (i.e. in rows, with the same number of rows as columns).

Just before the 200,007 Crucibilis Final, in which Dave Stevis beat Henry Stevens to win a record 7,000th Crucibilis Title, the game was changed slightly to an old version that had been invented by one of the greats of 100,000 years previously, Dave Joseph. In Joseph's variation, Nookers Plus, the reds were rearranged (without changing the total number of reds) so that they formed a rectangle with more rows than columns. The idea was that this would make the game more difficult and more interesting.

Joseph's original plan had been to make the number of rows 1,000 more than in the old version of Nookers, but this was not possible. It was found that there were, in fact, exactly 991 more rows of reds in Nookers Plus than there had been in Nookers.

How many balls are there on the table at the start of a game of Nookers?

[Semih_Sayginer]

According to my page, my average daily activity is 27½ posts.

I don't recall posting half a message!

perhaps half of one was removed by a bbc moderator who had broke in here during their closing time?

[Robert602]

Hidden answer follows to Q4, select the text below to reveal it:
(962537588114?)
I'll type up my method (to show I didn't use excel!) if it's confirmed.

[Robert602]

Possible answer follows to Q3, hidden again.

(Ask the first witch:
"What would either of the other witches say if asked what the remaining witch would say when asked the question 'Is it one of the first four paths'?"

The answer will be truthful whichever witch is asked, because the answer has been reversed by one lying witch and reversed back again by the other one before being reported to you.

The second and third witches can be asked almost the same question with a variation on the last part, halving the number of possible paths each time until it's down to one possibility.
)

[Robert602]

DGE, about question 2. Do we know beforehand that the weights we're measuring are exact whole ounces? Eg, if we can show that something is heavier than 1oz, but lighter than 3oz, can we assume it's 2oz without needing to balance the scales?

Edit: Never mind, got it, answer below:
(1,3,9,27)

[davis_greatest]

Robert602 is off to a flying start

Robert602, I like the idea of writing in hidden text! I'll have to learn how to do that. Is it easy?

Your answers to questions 2 and 3 are spot on. (And regarding your query on question 2 before you solved it - you MUST be able to make the scales balance exactly for every possible weight that is a whole number of ounces from 1 to 40.)

Your answer to question 4 seems astonishingly close without being correct. I am wondering if you made a small slip in the arithmetic... I shall wait for the correct answer before putting up question 5!

[Robert602]

Hmm. Checked my arithmetic and I don't think I've slipped up there, might be wrong in the method, depends how close I am really?

I am including all of the balls on the table, which as I understood it is reds + 2 each of the 6 colours and 2 whites.

As for hiding text, I just know how annoying it is when you see one of these puzzles and accidentally read the answer before getting a chance to attempt it. Some forums have spoiler tags to hide text, but on here I had to make do with setting the text colour to the same colour as the background.

Pretty easy to do, just put the text in tags like this, but use square brackets instead of curly ones:
{COLOR=#f1f1f1} Hidden text here {/COLOR}
My method:
(
Let x be the number of rows in the original square-packed Nookers. The number of balls is therefore x².
The number of rows in Nookers plus is then x+991 as given, and the number of columns is x-a, where a is unknown. Equating number of balls:
x²=(x+991)(x-a)
x²=x²+x(991-a)-991a
991a / (991-a) = x
x must be an integer, so 991-a must divide 991a exactly. The most obvious way to do this (though probably not the only way?) is to make the denominator 1 by setting a=990. Then, x=991*990 so the number of reds is =
(991*990)², or 962537588100.

I actually half expected there to be two ways of doing it and one of them to be ruled out by the 'x+1000 is not possible' bit, but alas.

Great questions though.)