Puzzles with numbers and things

[davis_greatest]

Question 9 - Pot Smack

The eagerly awaited new snooker tournament, Pot Smack, takes place amid much fanfare. 8 players are invited to play, and each is given a random seeding.

The format is a knock-out, with each match being the best of one frame. There are three small subtleties, however, at least two of which differentiate the games from normal snooker:

1) In Pot Smack, each time that a player makes a pot, he must go and smack one of the other 7 competitors.

It happens that the only players ever to get smacked are Maun Shirty and Damon Grott.
Other than Shirty and Grott, the other 6 players collectively give Shirty and Grott an equal number of smacks each during the tournament. Shirty, however, always smacks Grott. Grott always smacks Shirty.

2) Because the matches are so short, to keep the crowd entertained with lots of pots and smacks, players are required to continue a frame until all the balls (including the final black) are potted. After potting the final black, the next smack must be particularly hard on the face of Grott or Shirty.

3) The use of illegal substances such as pot or smack is strictly prohibited at all times.


At the end of the tournament (which had no re-racks or fouls throughout), won by the higher seeded of the two finalists, I notice something remarkable:

a) the number of smacks that Grott received during the tournament was equal to Shirty's seeding to the power of Grott's seeding; and
b) the number of smacks that Shirty received during the tournament was equal to Grott's seeding to the power of Shirty's seeding.

The tournament winner's face was unsmacked throughout the tournament, but he was given a special smack when he collected the trophy.

Questions:

i) What are Shirty's and Grott's seedings?
ii) State whether it is possible, from the information given, to determine the seedings of the winner and the runner-up of the tournament.
iii) If it is possible, what the seedings of the winner and the runner-up of the tournament?

(Note: You must prove that (or explain why) your answers are right!)

[davis_greatest]

doing the dance
and thanks to you for taking care of the explaining part (that isn't my strength)

I noticed that a pair of 'mirror#s' (sorry, I like symmetry) add up to 1.111.111.110, multiplied that # by the possible # of permutations devided by 2 (as we are talking pairs before)

Well, yes, your "pairs" would add up to that, because 1,111,111,110 is just 555,555,555 x 2

[snookersfun]

doing the dance
and thanks to you for taking care of the explaining part (that isn't my strength)

I noticed that a pair of 'mirror#s' (sorry, I like symmetry) add up to 1.111.111.110, multiplied that # by the possible # of permutations devided by 2 (as we are talking pairs before)

[Semih_Sayginer]

....and here was me trying to throw in a couple of wrong answers before i posted the correct one, so i didnt look too smart. erm, yeah, right.

well done!

im off to do the easy su doku in THE SUN

[davis_greatest]

Congratulations, snookerfun!

24.999.999.998????
or rather 319.999.999.997?
now reached about 2X10^14!
OK, my final #: 201.599.999.798.400 (am I anywhere in the neighborhood at all???)

Congratulations, snookerfun! Your last answer of 201,599,999,798,400 is correct!

For those who tried, here are a couple of quick ways of getting to the answer:

(1) If you write out all the 9-figure numbers, for which the digits 1, 2, 3, ..., 9 each appear once, then the digits 1 to 9 will each appear, of course, with equal frequency in each location. Therefore, the average 9-figure number in the list is 555,555,555 (because 5 is the average of 1,2,3,4,5,6,7,8,9).

How many numbers are there? Well, there are 9 possible figures that can appear as the first digit, leaving any of 8 figures that can then appear as the 2nd digit, leaving any of 7 figures than can then appear as the 3rd digit, ..., leaving only one possible figure to appear as the last digit.

So there are 9x8x7x6x5x4x3x2x1 (written "9!" or pronounced "9 factorial") = 362,880 numbers.

So the answer is 362,880 x 555,555,555
= 362,880 x 5 x 111,111,111
= 1,814,400 x 111,111,111 (which is easy to write out on paper or multiply with a computer)
= 201,599,999,798,400


(2) Another way:

Look at the units column. How many numbers appear with 1 in the units column? It is the number of ways of arranging the remaining 8 digits in the other 8 places, i.e. 8x7x6x5x4x3x2x1 (written "8!" or pronounced "8 factorial") = 40,320.

This is the same as the number of ways of 2 appearing in the units column, or 3, ..., or 9, so add up the units in all the numbers and you will get

8! x (1+2+3+4+5+6+7+8+9) = 40,320 x 45 = 1,814,400.

Now, add up the "tens" column, and you will get 1,814,400 x 10

Add up the "hundreds" column, and you will get 1,814,400 x 100

etc

So the total is

1,814,400 x 111,111,111 (as before)
= 201,599,999,798,400

[Semih_Sayginer]

1937102445

?

[snookersfun]

24.999.999.998????
or rather 319.999.999.997?
now reached about 2X10^14!
OK, my final #: 201.599.999.798.400 (am I anywhere in the neighborhood at all???)

[davis_greatest]

146966400

? (I am useless at math, I know)

More than that. Sorry!

[April madness]

146966400

? (I am useless at math, I know)

[davis_greatest]

,,,362,880 ?

Afraid not that either.

[Cessy143]

,,,362,880 ?

[davis_greatest]

17433922005

?

Afraid not. Sorry.

[Semih_Sayginer]

So, question 8 - apologies if this is too easy... I wanted to keep things moving:

What is the sum of all the nine digit numbers in which each of the digits 1, 2, 3, .... ,9 appears exactly once?

17433922005

?

[davis_greatest]

Question 8 - Adding up

So, question 8 - apologies if this is too easy... I wanted to keep things moving:

What is the sum of all the nine digit numbers in which each of the digits 1, 2, 3, .... ,9 appears exactly once?

[snookersfun]

Excellent! Nice solution, Vidas. Well done.

Slightly different from, but equallly valid to, the way I was hinting it could be done with the title "Snooker Tricolore" and the colours of the walls, ceiling and door being red, white and blue. You can colour the floor in a regular pattern with these three colours.

(T=toilet, r=red, w=white, b=blue - unfortunately I can't make the columns line up very neatly)

T r w b r w b r
r w b r w b r w
w b r w b r w b
b r w b r w b r
r w b r w b r w
w b r w b r w b
b r w b r w b r
r w b r w b r w

Each patch of cloth has to cover one red square, one white square, and one blue square. However, as there are 22 reds, 21 whites and 20 blues, rather than an equal number of each, you can also see that the covering cannot be done.

This is what I meant much better explained, I have to admit.

if you devide the original 64 squares into three types of sqares (in sequence),...

just if you had stuck the toilet into upper right hand corner you would have gotten even # of different squares, so therefore I needed my symmetry statement...