Puzzles with numbers and things

[davis_greatest]

I'm afraid there will be no kisses from Gordon, at least not in this tournament.

Will you explain?

[abextra]

OK, i'll try.

In total, there are 4x147147+2 rounds to play and 2^(4x147147+2) players to enter the competition. (I think, I will write x instead of 147147 and y instead of 2^x, just to make my life easier).
Then there will be 2^(4x+2) = 4(y^4) players i.e.

4(y^4) bronze medals,
2(y^2) silver medals,
2y gold medals and
2(y^2)+2y+1 medals in the first row.

4(y^4)+z (the number of iridium medals) must be divisible by the number of medals in the first row without a reminder, but I really didn't know, how to do it, so I counted on fingers and found out, that one iridium medal is enough to complete the last row. This is pretty nice formula

4(y^4)+1=(2y^2+2y+1)(2y^2-2y+1),

maybe I should know this one from school. It also means, that there will be
2y^2-2y+2 rows in total.

So, there will be only one iridium medal, it will go to the champion and as the champion won't lose a match, Gordon can save all his kisses for you, davis_greatest!

[abextra]

Your pets are extremely cute, davis_greatest, but aren't they too little to travel alone and work so hard?

[davis_greatest]

This is pretty much a perfect answer - very elegant. Well done! The interesting thing here is that the “147147” number could be any number at all – Gordon doesn’t have to kiss anybody!

...maybe I should know this one from school. It also means, that there will be
2y^2-2y+2 rows in total.

This was the only slip... there are in fact 2y^2-2y+1 rows in total - otherwise spot on.

So, there will be only one iridium medal, it will go to the champion and as the champion won't lose a match, Gordon can save all his kisses for you, davis_greatest!

Yes, he can!

I'll fill in the gaps but your answer was nicely written...

Let x = 147147.

The winner gets a platinum medal, so number of platinum medals = 1

To reach the final after winning a gold medal, a player must win a further x=147147 matches.
So number of gold medals = 2^(x+1)

To win a gold medal after winning a silver medal, a player must win a further x=147147 matches.
So number of silver medals = 2^(2x+1)

A player who wins a silver medal is at the halfway stage and needs to win a further 2x+1 matches to win the tournament (2x to reach the final plus one to win it).
Therefore, to win the tournament from the first round, a player must win 2.(2x+1) = 4x+2 matches

So number of bronze medals = 2^(4x+2)

Number of silver + gold + platinum medals = 2^(2x+1) + 2^(x+1) + 1

Let number of iridium medals to be determined, be Z

Number of bronze + iridium medals = 2^(4x+2) + Z

For a perfect rectangle to be formed, 2^(4x+2) + Z must be divisible by [2^(2x+1) + 2^(x+1) + 1].

So …. you need to divide 2^(4x+2) by [2^(2x+1) + 2^(x+1) + 1].

Now,

2^(4x+2) + 1 = [2^(2x+1) + 2^(x+1) + 1] x [2^(2x+1) - 2^(x+1) + 1]

(i.e. doing the division gives you [2^(2x+1) - 2^(x+1) + 1] with a remainder of -1)

So Z = 1

Therefore, one person gets an iridium medal. He must therefore be the winner of the tournament and therefore loses no matches.

So Gordon doesn’t have to kiss anyone! What a shame! The answer to the question is therefore zero.

Scoreboard to follow...

[davis_greatest]

Your pets are extremely cute, davis_greatest, but aren't they too little to travel alone and work so hard?

They enjoy it!

[davis_greatest]

abextra, the formula you give 4y^4 + 1 = (2y^2 + 2y + 1)(2y^2 - 2y + 1) has an interesting bit of history behind it.

In 1869, the mathematician Landry announced the factorisation of 2^58 + 1. He found that

2^58 + 1 = 5 x 107367629 x 536903681 and said:

"No one of the numerous factorisations of 2^n +/-1 gave as much trouble and labour as that of 2^58 + 1 ... if we lose this result, we shall miss the patience and courage to repeat all calculations that we have made and it is possible that many years will pass before someone else will discover the factorisation of 2^58 + 1."

In fact, it was easy! Simply write:

2^58 + 1 = (2^29 + 2^15 + 1)(2^29 - 2^15 + 1)

This is a special case, where x=14 of:

2^(4x+2) + 1 = [2^(2x+1) + 2^(x+1) + 1][2^(2x+1) - 2^(x+1) + 1]

or, putting y=2^x:

4y^4 + 1 = (2y^2 + 2y + 1)(2y^2 - 2y + 1).

So the factorisation done by Landry, of which he was so proud, could have been done in minutes, not years!

This was first shown by Eduard Lucas, who invented the Tower of Hanoi puzzle.

[davis_greatest]

Congratulations abextra! Closing the gap in the race for 2nd place...

SO HERE IS THE SCOREBOARD AFTER ROUND 41:

snookersfun……………………….…..18½
Vidas……………………………………….8½
abextra……………………………..…...7½
robert602…………………………………5
davis_greatest…………………..……5
elvaago...............................2
Semih_Sayginer.....................½

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[davis_greatest]

Round 42 - who won the thrilling decider?

I am watching the Grand Prix final on Astronomicus, after all those very, very many rounds have been played. It is a best-of-17-frames match, and Robotsdaughter has just taken a 4-1 lead against Cantcope.

Unfortunately, I have to stop watching, but I know that Cantcope will have a 52% chance of winning each future frame (and Robotsdaughter a 48% chance), and that each frame will be independent of all the others.

The following day, my friend asks me "Did you see the final last night? It was a thrilling deciding frame. 9-8, on the black!"

Who is more likely to have won the match? And what is the chance that that player won?

[abextra]

If Cantcope has a 52% chance to win every future frame, then he should have a 52% chance to win the final frame and he is more likely to have won the match? And why am I sure, that this is not this simple?

[davis_greatest]

Yes, as easy as it seemed!

If Cantcope has a 52% chance to win every future frame, then he should have a 52% chance to win the final frame and he is more likely to have won the match? And why am I sure, that this is not this simple?

Yes, it is that simple - a trick question! Given that the match went to a final frame, Cantcope had a 52% chance of having won it and so was more likely to have won.

If we didn't know that the match had gone to a final frame, but only knew that Robotsdaughter was leading 4-1, then it would in fact have been more likely that Robotsdaughter had won (by any score) - in fact, with a chance of about 76.52% to Cantcope's 23.48%. But the fact that we know that the scores reached 8-8 - and then went to a decider - changes everything.



SO HERE IS THE SCOREBOARD AFTER ROUND 42:

snookersfun……………………….…..18½
Vidas……………………………………….8½
abextra……………………………..…...8½
robert602…………………………………5
davis_greatest…………………..……5
elvaago...............................2
Semih_Sayginer.....................½

[abextra]

I like this kind of tricks, davis_greatest! How big is the chance you are going to ask this type of questions in the future? Thank you for the story (#546) and points! I've had some luck today!

[davis_greatest]

Round 43 - Sweet 16

In the scenario of round 42 (message 548) who is more likely to have won the penultimate (16th) frame? What is the chance that that player did so?

[Semih_Sayginer]

In the scenario of round 42 (message 548) who is more likely to have won the penultimate (16th) frame? What is the chance that that player did so?

surely:-

cantcope must win the penultimate (16th) frame or hed be beat 7-9

and.....

his chance would have been a 52% chance as stated

?

[davis_greatest]

surely:-

cantcope must win the penultimate (16th) frame or hed be beat 7-9

Afraid not. It is perfectly plausible that Cantcope took the lead and that Robotsdaughter then levelled.

and.....

his chance would have been a 52% chance as stated

?

Isn't the first part of your answer saying that Cantcope MUST have won the 16th frame (so chance 100%), and the 2nd part then saying instead that it was a 52% chance?

[Semih_Sayginer]

yeah, so it wasnt another trick

just kick me next time. i dont mind.

so, we actually have to work something out? ill give that a miss then.