The worlds first riddle!

5^2*sqrt(4)+4/4=51

Rap

did i say u cud do that rap

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]

i'm rapturous, but my fave solution is:

(4^4 - 52)/4

did i say solution

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

DKEST

KMLAN

As a linkman, I can try to enlighten you about finance, being the kindest poster here..

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

Francis takes finance, what an arbitrage.

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.

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.

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

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

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

Hiya guys; in a terrible rush - just time to say: Rap's answer is just as good as it gets.

BBL.

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

Good Gravy no...

Yes, but I don't know how to show it on a computer

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

this thred makes my brain hurt...

Thanks rap, now I know how to show what I got.

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

Hi guys;

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




Rap:

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.


Cool!



TTH:

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

I never doubted you for a second!

"…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!



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!