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The worlds first riddle!

 
 
raprap
 
  2  
Reply Mon 7 Jul, 2008 08:56 pm
5^2*sqrt(4)+4/4=51

Rap
0 Replies
 
solipsister
 
  2  
Reply Mon 7 Jul, 2008 09:46 pm
did i say u cud do that rap
0 Replies
 
markr
 
  2  
Reply Tue 8 Jul, 2008 01:31 am
ROWS OF TENNIS BALLS
These procedures will fill the rows a column at a time.

[size=7]k is odd (k = 2n+1):
Column 1 consists of the numbers 1 to k in that order from top to bottom.
Column 2: Starting with row n+2 and working down (wrap back to the top when you get to the bottom), enter the numbers k+1 to 2*k.
Column 3: Starting with the middle row and working down (wrap back to the top when you get to the bottom), enter the numbers 2*k+1 to 3*k, but skip n rows between consecutive numbers.

k is even (k = 2n+2):
Column 1 consists of the numbers 1 to k in that order from top to bottom.
Column 2: Place the number k+1 in row 1. Starting with row n+3 and working down (wrap back to row 2 when you get to the bottom), enter the numbers k+2 to 2*k.
Column 3: Place the number 3*k+n+1 in row 1. Starting with row n+2 and working down (wrap back to row 2 when you get to the bottom), enter the numbers 2*k+1 to 3*k-1, but skip n rows between consecutive numbers.
[/size]
0 Replies
 
TTH
 
  2  
Reply Tue 8 Jul, 2008 09:59 am
solipsister wrote:
did i say u cud do that rap
Laughing Laughing
0 Replies
 
solipsister
 
  2  
Reply Tue 8 Jul, 2008 09:18 pm
i'm rapturous, but my fave solution is:

(4^4 - 52)/4

did i say solution
0 Replies
 
Tryagain
 
  1  
Reply Wed 9 Jul, 2008 11:18 am
Hi-Di-Hi Folksies, long time no see:

Aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it deosn't mttaer in what oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist and lsat ltteer be at the rghit pclae. The rset can be a total mses and you can sitll raed it wouthit porbelm. This is bcuseae the huamn mnid deos not raed ervey lteter by istlef, but the word as a wlohe.



ROWS OF TENNIS BALLS
Mark, what a great answer - I don't have a clue.

Slippy, you remind me of someone; it will come to me!




AFNCE Question

DKEST Question

KMLAN Question
0 Replies
 
Francis
 
  2  
Reply Wed 9 Jul, 2008 11:28 am
As a linkman, I can try to enlighten you about finance, being the kindest poster here.. Very Happy
0 Replies
 
solipsister
 
  2  
Reply Wed 9 Jul, 2008 08:54 pm
Tryagain wrote:


Slippy, you remind me of someone; it will come to me!




Perhaps we met one summer, knowing your proclivity to enjoin us in maths questions.

Francis takes finance, what an arbitrage.
0 Replies
 
Tryagain
 
  1  
Reply Thu 10 Jul, 2008 07:15 am
Francis; as ever your replies eclipse the questions.



Dear Slippy, now I remember; it was at the Christmas party, I was boring you under the mistletoe; the memory still brings a smile to my face.
0 Replies
 
jove
 
  2  
Reply Thu 10 Jul, 2008 07:56 am
Hello Try,,, i miss you (i'm a wimpy turd). can anyone see a likeness of a man in the gem on the right and a female body in the gem on the left?,,, take your time ~

http://www.gemfix.com/images/stones/ruby/ruby60.jpg

Zion Bible College (from Rhode Island) is moving into Bradford College of Ma. this Sept. after repairs are done ~ if anyone is interested in working or attending a Bible College in Bradford, Ma.

The grounds are pretty and serene. There is a small pond that has duck's that visit near the quaint and charming New Englandish, curved, bridge. It's beautiful there in the fall.
0 Replies
 
Tryagain
 
  1  
Reply Sat 12 Jul, 2008 11:16 am
Dearest Jove etc, berate yourself not; to us you are a princess. I can see the images in the gems; do I win a prize? I will have to skip bible college this year due to flunking the practical test.



The A2KLeague Baseball Series ended last Thursday night with a come-from-behind victory by the Redneck Riddlers over the Trivia Twits.

The series is a best-of-seven format, where the first team to win four of seven games is the champion. After the first six games the series was tied at three games apiece.

What is the probability of that happening Question

More precisely, assume that two evenly matched teams play each other in a best-of-seven series. What is the probability that all seven games have to be played before the series is decided Question
0 Replies
 
raprap
 
  2  
Reply Sat 12 Jul, 2008 03:24 pm
If the teams are perfectly matched, the probabality of team winning is the same as losing. So the probability of any occurrance is effectively the same---the probability, then, of a particulat situation is then goverened by the combinations, which is =g!/((g-w)!w!))

Therefore tied after 6 games is
6!/(3!*3!)*(1/2)^6=6*5*4/(3*2*1)*1/2^6=20/64=10/32=5/16
Or the probability to be tied after six ganes is 5/16
BTW to be tied after six games means there will be a seventh game and since this is the only way to a seventh game--the probability is the same.

Rap
0 Replies
 
Tryagain
 
  1  
Reply Sun 13 Jul, 2008 11:06 am
Hiya guys; in a terrible rush - just time to say: Rap's answer is just as good as it gets. Cool Cool

BBL.
0 Replies
 
Tryagain
 
  1  
Reply Sun 13 Jul, 2008 05:50 pm
As promised I return with a cracker of a puzzle.

A six-pointed star is drawn with six lines and twelve vertices. (Join six triangles)

Can you arrange the integers 1 through 12, one on each vertex, so that the four integers on each line add to 26 Question
0 Replies
 
mismi
 
  2  
Reply Mon 14 Jul, 2008 04:29 pm
Good Gravy no...
0 Replies
 
TTH
 
  2  
Reply Tue 15 Jul, 2008 07:30 pm
Tryagain wrote:
A six-pointed star is drawn with six lines and twelve vertices. (Join six triangles)

Can you arrange the integers 1 through 12, one on each vertex, so that the four integers on each line add to 26 Question
Yes, but I don't know how to show it on a computer Sad
0 Replies
 
raprap
 
  2  
Reply Wed 16 Jul, 2008 10:14 pm
After awhile I got this one

.......1
.5 .6 12 .3
..10.. 11
.9.. 7.. 8. 2
.....4


But it isn't unique

The method is finding all permutations of four integers less than 13 and greater than 0 that sum to 26, and then setting these permutations into the star of david pattern to find the ones that fit. Since there are 32 permutations there are more than one pattern.

Rap
0 Replies
 
Rockhead
 
  2  
Reply Wed 16 Jul, 2008 10:26 pm
this thred makes my brain hurt...

Shocked
0 Replies
 
TTH
 
  2  
Reply Thu 17 Jul, 2008 07:43 am
Thanks rap, now I know how to show what I got.

........1
9.10..2.5
....7.....12
..8..4..3..11
........6
0 Replies
 
Tryagain
 
  1  
Reply Thu 17 Jul, 2008 01:56 pm
Hi guys;

Miss mi; you were close…but don't give up the day job! Laughing




Rap:

After awhile I got this one

.......1
.5 .6 12 .3
..10.. 11
.9.. 7.. 8. 2
.....4 Cool Cool


But it isn't unique Cool

The method is finding all permutations of four integers less than 13 and greater than 0 that sum to 26, and then setting these permutations into the star of david pattern to find the ones that fit. Since there are 32 permutations there are more than one pattern.


Cool!



TTH:

........1
9.10..2.5
....7.....12
..8..4..3..11
........6 Cool Cool

I never doubted you for a second! Very Happy

"…now I know how to show what I got."

Oh boy! Let me get a front row seat and popcorn before you let them puppies out! Razz



Starting from the top clockwise:

12, 6, 9, 5, 3, 11, 1, 10, 2, 8, 7, 4.

Which gives:

12, 6, 5, 3.
9, 5, 11, 1.
3, 11, 10, 2.
1, 10, 8, 7.
2, 8, 4, 12.
7, 4, 6, 9.



One program repeatedly guessed six different integer values for the six vertices in the middle, then computed the other six from the solution of a system of linear equations.

The integer at an outer vertex x is computed as x=(26+a+b-(c+d+e+f))/2, where c,d,e,f are the inner vertices closest to x, and a,b are the inner vertices farthest from x.

7 = (26 + 5+11 - (4+6+8+10)) /2
2 = (26 + 6+5 - (4+8+10+11))/2
It succeeded when the resulting twelve integers were exactly (1,2,...,12) in some order.

Another program used "hill climbing" techniques: place the twelve integers (1,2,...,12) on the vertices randomly, then repeatedly find and execute a "beneficial trade": a pair of integers such that when their positions are traded, some "figure of merit" is improved. In this case the figure of merit was the mean square error of the line sums, namely the sum, over the six lines, of the squares of the differences between their current line sum and the desired line sum of 26.


1. Find all the combinations of 4 of the first 12 integers which sum to 26. Of the 495 possible (12 items taken 4 at a time), 33 sum to 26.

2. For the set of 33, it then generated all the groups of 6 where each integer is used exactly twice. There are 514.

3. It then searched those 514 for the cases where there is only one (or none) shared values been every pair of 6 sums. There are 20. I then randomly picked one of the 20, and in a few minutes (manually) juggled the placement on the star.



Rock wrote (try saying that with a mouth full of gumbo), "this thred makes my brain hurt..."

Welcome to the club! Laughing
0 Replies
 
 

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