Puzzles with numbers and things

[davis_greatest]

Most numbers have an even number of factors, because factors "work in pairs": ex 12 = 1*12 or 2*6 or 3*4, so factors of 12 are 1,2,3,4 6 and 12.
Squares however have an odd number of factors: ex 16 = 1*16, 2*8, 4*4
so factors of 16 are 1,2,4 8 and 16. The "odd" results from one of the pairs featuring twice the same factor, the square root.

Now if n is the order of a ball in the row (n=5 for the fifth ball f.i.) then this ball is rolled from one side to the other once for each factor of n. Therefore all balls with n = "a square" end up on the right, and all other balls on the left.

As we have 10000 balls and 10000 is 100*100, we have 100 "square" balls (one for each value k*k for k = 1 to 100).
Therefore we have 100 balls on the right and 9900 balls on the left.

Hope this is understandable enough ...

Round 161

Thanks, Monique - nicely explained.

Indeed, if we number all the balls from 1 to 10,000, the squares (1, 4, 9...) end up on the right cushion, as they have an odd number of factors and so are rolled an odd number of times. All the other balls have an even number of factors, and so end up on the left as they are rolled an even number of times.

For any n, if k is a factor of n, then n/k is also a factor, so the factors come in pairs. The only exception is where n is a square, say n=k²; then k and n/k are the same.


Round 163

OK, snookersfun and Monique (Edit: I am being kind here ) have solved round 163. Time for someone to post an answer please!

[Lee Vilenski]

Ah, yes, I see that now... Just the language I think..

[Monique]

R161 Roll across

Most numbers have an even number of factors, because factors "work in pairs": ex 12 = 1*12 or 2*6 or 3*4, so factors of 12 are 1,2,3,4 6 and 12.
Squares however have an odd number of factors: ex 16 = 1*16, 2*8, 4*4
so factors of 16 are 1,2,4 8 and 16. The "odd" results from one of the pairs featuring twice the same factor, the square root.

Now if n is the order of a ball in the row (n=5 for the fifth ball f.i.) then this ball is rolled from one side to the other once for each factor of n. Therefore all balls with n = "a square" end up on the right, and all other balls on the left.

As we have 10000 balls and 10000 is 100*100, we have 100 "square" balls (one for each value k*k for k = 1 to 100).
Therefore we have 100 balls on the right and 9900 balls on the left.

Hope this is understandable enough ...

[davis_greatest]

Easier (partly as this has similarities to another question I posted during the previous 160 rounds)

Round 161 - Roll across

I have lined up 10,000 balls in a long line down the left hand side of my incredibly long snooker table.

Then I roll every ball to the other side (i.e. in this case, from the left to the right hand side), always keeping each ball the same distance from the top and bottom cushions.

I then roll every 2nd ball from whichever side it is on to the other side (i.e. the 2nd, 4th, 6th ball etc from the top cushion). So, in this case I will roll balls 2, 4, 6, 8 etc from right to left.

I then roll every 3rd ball from whichever side it is on to the other (i.e. the 3rd, 6th, 9th etc from the top cushion).

I keep doing this - every 4th ball, every 5th, 6th, ... swapping the side of the table they are on, by rolling them across....until I have rolled ever 10,000th ball (there is only one of those!).

How many balls finish on each side?

OK, just remembered that round 161 hasn't been closed. snookersfun and Monique have sent me solutions - someone please post the answer on this thread!


And, I did promise I'd try to think of something else, so...

Round 163 - Weighty problem

I have a huge collection of red snooker balls, all pretty standard (not especially light or heavy for snooker balls) - and in fact they all weigh exactly the same (each ball is a whole number of grams exactly).

I divide my massive collection into two equal groups, and find that I can make a "triangle" of balls with each group... except that each "triangle" has one ball missing at the apex. (For example, if the "triangle" were the same size as a normal triangle of reds in snooker, there would be 14 reds instead of 15, as the first red at the apex would be missing.)

So, I have two identical "triangles" of balls (each triangle having one ball missing).

Bored with this, I then scoop up all the balls (combining both groups again), and then put them all in rows to form a very big rectangle on my exceptionally large and exceptionally sturdy snooker table. I make the rectangle as wide as possible, except that it must be at least as long as it is wide.

Even more bored, I finally scoop up one row of balls (along the width of the rectangle), put them on my huge weighing scale and see how much the row of balls weighs. I am pretty chuffed to see that it is about 147kg .... well, 147.899kg to be exact.

How many rows of balls are left in the rectangle, after I had scooped up that first row?

[davis_greatest]

Lagermike has not said that he is 72.5% certain that Paul is right 75% of the time....

Yes, he has for the purposes of this puzzle - it says so above! I know that the real lagermike did not say that on the other thread, but what he said did not readily lend itself to a simple puzzle, so poetic licence was required and accepted.

And, as Monique found and The Statman also implied, 61.25% is the correct answer! This being...

Chance that PaulTheSoave is right

= Chance that lagermike is right x Chance that PaulTheSoave is right given that lagermike is right
+
Chance that lagermike is wrong x Chance that PaulTheSoave is right given that lagermike is wrong

= 72.5% x 75% + 27.5% x 25%
= 61.25%


And yes, hopefully next question to follow after I've had lunch (if I think of anything)...

[Monique]

Make of that what you will.

I'm afraid D_G will come up with yet another puzzle

[The Statman]

and if you want an easier one, try this

Round 162 - Certainly not!

PaulTheSoave believes that he was present at a match sitting near the baulk side on row H, with Selby making a break against Murphy. lagermike believes that PaulTheSoave is right 75% of the time (although lagermike is only 72.5% certain of this).

Whenever lagermike is wrong, the opposite applies (in this case, if lagermike were wrong, PaulTheSoave would be wrong 75% of the time).

What are the chances that PaulTheSoave was present at a match sitting near the baulk side on row H, with Selby making a break against Murphy?

Firstly, it would be useful for PaulTheSoave to refer to his tickets (or enquire of the box office if he has binned them) and ascertain for which sessions he was in Row H. This would have an approximately 38.7% chance of confirming that Paul is wrong (i.e. if he was not seated in Row H for any session in which Selby was playing Murphy).

Lagermike has not said that he is 72.5% certain that Paul is right 75% of the time. He has stated merely that he is 72.5% certain that Paul is 100% right on this one occasion.

The (hitherto unpublished) fact that I was thinking of Selby as I wrote my initial piece (although only 45.7% certain, hence the hitherto unpublishedness) does add weight to Paul's suggestion.

Indeed, since it is a fact of which Paul is 75% and Mike 72.5% certain, this means that, overall, the two of them are 73¾% certain that indeed it was Selby.

However, with Mike's uncertianty and Paul's uncertainty, not to mention my initial even greater uncertainty, the one thing that we can be certain of is that there is certainly a fair amount of uncertainty about.

I note from Globalsnookercentre that, in frame 20 of the Selby-Murphy match, Selby scored a point, then a 31 break and then a 25 using eleven reds (possibly ten, depending on whether Murphy's six points arose from a break or a foul). Thinking about it, that might not be a low enough average-per-red.

Make of that what you will.

[davis_greatest]

161, Well all the prime numbers would stay on the right hand Cussion. But after that I got confused.

Sadly not. It's not really about prime numbers. In fact, all the prime numbers will finish on the left cushion. For example, think of ball 2, that gets rolled back to the left cushion and then never moves. The same for ball 3! ...

[Lee Vilenski]

161, Well all the prime numbers would stay on the right hand Cussion. But after that I got confused.

[davis_greatest]

Monique is first in with a quick answer to round 162. Next one to answer, please do so on the thread.

And snookersfun and Monique both got the answer to round 161 pretty quickly.... Any other takers?

[davis_greatest]

and if you want an easier one, try this

Round 162 - Certainly not!

PaulTheSoave believes that he was present at a match sitting near the baulk side on row H, with Selby making a break against Murphy. lagermike believes that PaulTheSoave is right 75% of the time (although lagermike is only 72.5% certain of this).

Whenever lagermike is wrong, the opposite applies (in this case, if lagermike were wrong, PaulTheSoave would be wrong 75% of the time).

What are the chances that PaulTheSoave was present at a match sitting near the baulk side on row H, with Selby making a break against Murphy?

[davis_greatest]

lol, quite a nice one actually. You got an easy one for me?

Easier (partly as this has similarities to another question I posted during the previous 160 rounds)

Round 161 - Roll across

I have lined up 10,000 balls in a long line down the left hand side of my incredibly long snooker table.

Then I roll every ball to the other side (i.e. in this case, from the left to the right hand side), always keeping each ball the same distance from the top and bottom cushions.

I then roll every 2nd ball from whichever side it is on to the other side (i.e. the 2nd, 4th, 6th ball etc from the top cushion). So, in this case I will roll balls 2, 4, 6, 8 etc from right to left.

I then roll every 3rd ball from whichever side it is on to the other (i.e. the 3rd, 6th, 9th etc from the top cushion).

I keep doing this - every 4th ball, every 5th, 6th, ... swapping the side of the table they are on, by rolling them across....until I have rolled ever 10,000th ball (there is only one of those!).

How many balls finish on each side?

[snookersfun]

ahem!!!! genius at work

I knew I had it coming...

[davis_greatest]

here is the number with proper spaces inserted:

741 041 431 921 521 221 021 311 211 501 401 79 69 98 88 18 08 37 27 56 46 75 65 94 84 14 04 33 23 52 42 71 61 9 8 1

now reading that from right to left (reversed numbers and all) (! and I have to admit I initially got the result, without doing that , it will be the sequence of a maximum break in snooker

Yes, it is a 147 maximum in snooker, written backwards! Congratulations! The break made at every visit to the table by Willie Thorne, hence the title, which I thought would give it away. Congratulations also to Monique who also got it.

...(! and I have to admit I initially got the result, without doing that

And I'm sure snookersfun won't mind me posting her earlier explanation :

...Don't quite have the beginning sorted out yet as you'll see...

74104 143192 15212 2102 131 121 150 140 179 69 98 88 18 08 37 27 56 46 75 65 94 84 14 04 33 23 52 42 71 6…1 9 8 1

so this being your number, I inserted these breaks which show last digit decreasing, e.g. 22110099...(although there are these strange patterns of 4x 8 and 4, but that might all make sense, seeing that one loses first digits, moves from 9 to 1 in second to last digit or whatever...) second to last digit is easier seen, when ordering the numbers into two columns (couples-with same last digit). One column would show that second to last digit odd second column even and increasing (although following even number always one smaller, but with same last digit).
Do you understand me at all?
So, 71 needs 61 to follow and after that 7 and 6 would be followed by 9,8 respectively and 9 of course by one.

Now, just explain to me the beginning of that sequence ... and don't tell me, there is a deeper meaning behind all that....

I won't post here my response to that.

[Lee Vilenski]

lol, quite a nice one actually. You got an easy one for me?