TMC OPEN 3: Picnic Edition--May 31, 2009--Stratford, CT.--($500+ in prizes!)

[iamfishman]

THE MANA CLASH 3- PICNIC EDITION*

Hi all.

First the neccessary info(applies to both tournaments):
*”What is this Picnic Edition thing all about?”
To celebrate the warm weather, and that this is the weekend after memorial day festivities, their will be picnic tables and someone grilling outside the store.  In between matches, have a burger, take a rest at a picnic table with other magic players, enjoy the sun, throw a Frisbee around, and then come back in read and refreshed for another round of magic!

When: Sunday, May 31, 2009
Registration: 10:00 AM - 11:00 AM
Tournies Starts PROMPTLY at 11:00 AM(To avoid a prolonged tournament, this will be strictly enforced. Latecomers will be subject to a round 1 loss.)

Where:
Gamingetc Tournament Center and SuperStore
555 Lordship Boulevard (Exit 30 off I95)
Stratford CT 06615
1-800-380-1115 or 203-849-0186

Lodgings:
Ramada Inn
225 Lordship Boulevard
Stratford, Ct 06615
203-375-8866(Mention “Magic Tournament” to get a special room rate)

Dopeness Factor:
For each positive integer n, let An = [(n + 9)!]/[(n − 1)!]
Let k denote the smallest positive integer for which the rightmost nonzero digit of Ak is odd. The dopeness factor is that rightmost nonzero digit of Ak.

Pre-reg(and prize):
To pre-reg(it is free, you don't need to pay at the time you pre-reg, and their will be a prize for a lucky person who pre-regs for the event) please PM Iamfishman on themanadrain.com or email me at Iamfishman2000@hotmail.com

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Tourney 1: Vintage
Cost: $25 Entry Fee
Proxies: 10 Proxies + $1 per proxy(15 proxies maximum)
Starts PROMPTLY at 11:00 AM(To avoid a prolonged tournament, this will be strictly enforced. Latecomers may be subject to a round 1 loss.)
Based on 30 competitors:
1st Prize – 260 store credit or the store will buy back the credit for $200 in cash
2nd Prize – 130 store credit or the store will buy back the credit for $100 in cash
3rd-4th Prize – 65 store credit or the store will buy back the credit for $50 in cash
5th-8th Prize – 35 store credit or the store will buy back the credit for $25 in cash
Prizes will be increased as attendance does.

This event will follow the Universal Tournament Organizer Guidelines.
http://magiceternal.com/UTOG.html
And count toward Vintage Player Rankings
http://www.themanadrain.com/index.php?topic=37059.0

Tourney 2: Legacy
Cost: $25 Entry Fee
Proxies: None
DCI Sanctioned: YES
Starts PROMPTLY at 11:00 AM(To avoid a prolonged tournament, this will be strictly enforced. Latecomers may be subject to a round 1 loss.)
Based on 30 competitors:
1st Prize – 260 store credit or the store will buy back the credit for $200 in cash
2nd Prize – 130 store credit or the store will buy back the credit for $100 in cash
3rd-4th Prize – 65 store credit or the store will buy back the credit for $50 in cash
5th-8th Prize – 35 store credit or the store will buy back the credit for $25 in cash
Prizes will be increased as attendance does.

Free Single Elimination Side event for Vintage for 2 non-blue duals.
Free Single Elimination Side event for Legacy for 2 non-blue duals.

Vintage vs. Legacy Magic Trivia(Free with Prizes for all participants)

The store stocks a huge selection of singles and will be selling and buying all day.

Questions, comments, concerns? Please share them all here.

Cya there,
Ray

[mdenny]

Is the dopeness Factor, Dopeness = An for just the Vinage or for both? and if it is for just the vintage, then would playing in both make it 2(An) or An+1?

Sounds sweet i will try to make it for the Vintage atleast

[sorcutt]

Sounds great! 

In theory the picnic is a great idea. However, MTG kids are afraid of the sun; it's science.     

[tehmajickguy]

Damn you Ray, not only for making me play more magic but making me choose which format I wanna play. Count me in.

[Harlequin]

Dopeness Factor:
For each positive integer n, let An = [(n + 9)!]/[(n − 1)!]
Let k denote the smallest positive integer for which the rightmost nonzero digit of Ak is odd. The dopeness factor is that rightmost nonzero digit of Ak.

The answer is 1   ... Assuming you ment binary right?

I figgured out how to solove it, then I hand-cranked the first 100 or so numbers to test it.  Then I made a formula for it and tested about 1000 more.  Then I decided its programing time... I let it ran for a minute or so and it got to around 5k and I decided I should probably do "real work" for a bit.  On second thought, I'll let it run while I go down for coffee and see if it gives me an answer when I get back.

I let you know how it turns out. 

EDIT: up to 15k or so with no answer ... however there is a chance by starting and stopping  it a few times I actually messed up by tallies.  Oh well I'll have to start back from the begining.

[forcethewill]

im there and im winning it... both of them! at the same time!

[oneofchaos]

Sounds great! 

In theory the picnic is a great idea. However, MTG kids are afraid of the sun; it's science.     

Not everyone has shroud

I'm def in, I take finals up until May 13th.  I'll be back in connecticut to eat burgers and drain tezzerets.

[Nomad]

Dopeness Factor seems to be 9?
I got at k=49 (and let's remember that at Worlds and the Pro Tour k is only 48):
k=45  k+9 =54, factorial = 2.30844E+71 k-1=44, factorial = 2.65827E+54
Ratio of factorials = 86839771951295900 (=54*53*52*51*50*49*48*47*46*45)

D'oh.  It's at k=45 not 49.  Never mind.

[Gaagooch]

Ray, you are amazing...I don't know what life would be like without you...

[Sporkcore]

im there and im winning it... both of them! at the same time!

Now that's something I'd like to see.

[T00L]

Damn it Ray! Don't you know I'm trying to quit!

[Diakonov]

That is a challenging Dopeness Factor.  I can't seem to solve it.

Isn't it impossible to get an odd number for the smallest non-zero digit if there is a single even number anywhere in the product?  How do you get a factorial to not use evens when it starts off at 10?

[Harlequin]

Dopeness Factor seems to be 9?
I got at k=49 (and let's remember that at Worlds and the Pro Tour k is only 48):
k=45  k+9 =54, factorial = 2.30844E+71 k-1=44, factorial = 2.65827E+54
Ratio of factorials = 86839771951295900 (=54*53*52*51*50*49*48*47*46*45)
D'oh.  It's at k=45 not 49.  Never mind.

Nope that the first number excell starts truncating though.  The true 54*53*52*51*50*49*48*47*46*45 = 86839771951296000

Hint below

I found the number of K, using my program.  K is somewhere around 17k I don't remeber it off hand.  Tomorrow I'll work to solve for the actualy digit.

The important thing to try is prime factorization.  Start with something like A3 = 3*4*5*6*7*8*9*10*11*12.
 Now prime factor it to A3 = 3* 2*2* 5* 2*3* 7* 2*2*2* 3*3* 2*5* 11* 2*2*3
So .. 2^9 * 3^5 * 5^2 * 11 

If we moved to the next number in the sequence, we lose the number 3 and gain 13.  Meaning we go down one 3^1 and go up a 13^3.  then we move to the next number, losing 4 and gaining 14.  So we go down 2^2 and gain 2^1 and 7^1.  So... why is this important.

Well 10^x can rewritten as (2 x 5)^x  Try to use this to remove "right zeros" ...........
How does that help us?

[Nomad]

Dopeness Factor seems to be 9?
I got at k=49 (and let's remember that at Worlds and the Pro Tour k is only 48):
k=45  k+9 =54, factorial = 2.30844E+71 k-1=44, factorial = 2.65827E+54
Ratio of factorials = 86839771951295900 (=54*53*52*51*50*49*48*47*46*45)
D'oh.  It's at k=45 not 49.  Never mind.

Nope that the first number excell starts truncating though.  The true 54*53*52*51*50*49*48*47*46*45 = 86839771951296000

I did slightly have my doubts as to how many digits Excel would keep, I think it might have managed if I'd calculated as written.  So looks like basically we have to head somewhere where x=y, if it prime factors out to 2^x * 5^y * (odd numbers).  Not certain that's possible, is it?, with the advance of each "block" of 10 numbers, we add 2 more 5s and 5 more 2s.

[oneofchaos]

Damn it Ray! Don't you know I'm trying to quit!

We are the mana drain. Shuffler your deck, and present it to your opponent. We will add your tech and sideboard to our own. Your cardpool will adapt to service ours. Quitting is futile.

[Harlequin]

Dopeness Factor:
For each positive integer n, let An = [(n + 9)!]/[(n − 1)!]
Let k denote the smallest positive integer for which the rightmost nonzero digit of Ak is odd. The dopeness factor is that rightmost nonzero digit of Ak.

I think I solved it:

Hint 1: from post before

The important thing to try is prime factorization.  Start with something like A3 = 3*4*5*6*7*8*9*10*11*12.
 Now prime factor it to A3 = 3* 2*2* 5* 2*3* 7* 2*2*2* 3*3* 2*5* 11* 2*2*3
So .. 2^9 * 3^5 * 5^2 * 7 * 11 

If we moved to the next number in the sequence, we lose the number 3 and gain 13.  Meaning we go down one 3^1 and go up a 13^3.  then we move to the next number, losing 4 and gaining 14.  So we go down 2^2 and gain 2^1 and 7^1.  So... why is this important.

Well 10^x can rewritten as (2 x 5)^x  Try to use this to remove "right zeros" ...........
How does that help us?
Hint 2:

For A3 when we can remove (2x5)^2 because we have 5^2.  When we do we are left with:
2^7 * 3^5 * 7 * 11 = 2395008.  Notice that A3 = 239500800 = 2395008 * (2x5)^2
So what are we looking for, how do we define K in this way?
Hint 3:

2 is the only even prime.  We are able to remove powers of two from our prime factor list by pairing them with powers of 5.  Which means if we can find a value K that gives us a prime factor list that has the same number of powers of 2 and powers of 5, when we remove those powers of 2 and 5 we end up with only odd numbers.  So thats how we find K.  Then finding the dopeness factor is just a mater taking the remaining numbers (after dividing out our 2's and 5's) and multipling thier ones digits together.  We can ingore any other digits higher than the ones spot.
ANSWER:

At first I started hand grinding the problem.  Which isn't that hard.  I got to about N=100 in about 20 mins.  The method I used was to keep a running tally of 2's and 5's starting with
N = 6 is 8-twos and 2-fives.  Then I just looked at the number that was leaving and the number coming in.  So K=7 we lose 6 (-1 two) and gain 16 (+4 twos) which leaves us at 11-twos 2-fives.  This is pretty quick, If you can look at a 2 digit number and tell if its divisible by 2, 4, 8, 16 etc.  However I noticed I made a mistake when 40 entered (I counted it as 1-five and 1-two)  Which botched EVERY set going forward because I was just keeping a tally.

So next I make a function for =Factor2 and =Factor5 which would give me the exact numbers to add and subtract.  I copied that list down to about 1000 or so and gave up on that method.  I noticed that most values hovered around 10-11 twos and 2-3 fives.  Which led me to belive our number could be around 5^9 10 or 11.  So it was time to write a loop for it.

So next I basically just wrote one big loop to look at the new number in and the new number out, ran it through my factor2 and factor5 functions and did the same "tally" idea but in reliable variables.  I let that cook and came up with my anser:

K = 78117   [with 8-twos and 8-fives thanks to 78125 = 5^7] 

Prod-  .-2- -5-
78117   0  0
78118   1  0
78119   0  0
78120   3  1
78121   0  0
78122   1  0
78123   0  0
78124   2  0
78125   0  7
78126   1  0
--------------
 ...... 8  8

Next we need to carve out powers of 10 so we can be left with a number useful in solving for the dopeness factor so...

78117   0  0 => 78117 / 1 = 78117
78118   1  0 => 78118 / 2 = 39059
78119   0  0 => 78119 / 1 = 78119
78120   3  1 => 78120 / 40 = 1953
78121   0  0 => 78121 / 1 = 78121
78122   1  0 => 78122 / 2 = 39061
78123   0  0 => 78123 / 1 = 78123
78124   2  0 => 78124 / 4 = 19531
78125   0  7 => 78125 / 5^7 = 1
78126   1  0 => 78126 / 2 = 39063

So our dopeness factor = the ones digit of ( 7*9*9*3*1*1*1*3*1*1*3 ) = 15309

So 9 is the answer.  With K = 78117 and A78117 looks something like n...n,nnn,nnn,900,000,000

[Diakonov]

Nice work, Harlequin.  Very impressive.

[sorcutt]

Not everyone has shroud

Knocking over the sun-sphere = win.  Simpsons style of course.

[LordHomerCat]

My brain exploded.  That is some serious maths.

[TheBrassMan]

Menendian can blame it on proxies, and Marske can blame it on lazy American TOs... but I think we all know the reason Vintage is dying is too much math in Tournament Announcement threads.

also not enough Brass Man

[ashiXIII]

Menendian can blame it on proxies, and Marske can blame it on lazy American TOs... but I think we all know the reason Vintage is dying is too much math in Tournament Announcement threads.

also not enough Brass Man

if Ray ran a tourney every single weekend, i'd drive 3.5 hours for a tourney every single weekend.

In other words, the real reason Vintage magic is dying is because Ray doesn't run enough tourneys

[honestabe]

I'd like to try and make this one

[wethepeople]

Is this the same place as the one a couple summers back? I might be able to make it depending on who else is going.

[HungryHungryHeifer]

also not enough Brass Man

I lol'd. Literally. While at work and everything! It's true though. There has not been nearly enough Brass Man!

[Duncan]

I’m afraid I have to select every blank line to check for white text from now on…

[iamfishman]

This event is almost here.  Don't forget...it still isn't too late to pre reg.

[oneofchaos]

This event is almost here.  Don't forget...it still isn't too late to pre reg.

I won't be making an appearance thanks to work.  I'll get my friend to play my deck and wear the cowboy hat tho.

[Oath of Happy]

So, what happened? Is there a report somewhere?

[iamfishman]

So, what happened? Is there a report somewhere?

Between end of the year stuff at school, my stag, and consequently nursing myself back to health for the last few days, I haven't been able to post anything, but it should be up within 24-48 hours.

[oneofchaos]

So, what happened? Is there a report somewhere?

Between my stag, and consequently nursing myself back to health for the last few day.

Is the nursing back to health a direct cause of the stag party?