
Content Count
623 
Joined

Last visited

Days Won
8
Content Type
Profiles
Forums
Calendar
Gallery
Blogs
Posts posted by Pickett


well...
SpoilerO=0, M=1, Y=2, E=5, N=6, D=7, R=8, S=9
Now, I'm not quite sure what you are asking in your question...as substituting in, you would get this:
065 065+2 065+4 065+17 =?
I don't know what you want us to do with those...I guess you could get
65 67 69 82
Are you wanting those multiplied (24640590)? added (283)? concatenated (65676982)? resubstituted (NE ND NS RY)?

Depends...
SpoilerIf you go with purely "expected damage value" as your basis for "best", then it would be
 Shotgun
 Pistol
 Railgun / Grenade
 Rifle
 Knife
However, I personally would rather have something that guarantees SOME damage over something that has a 50% chance of doing 0 damage (so Railgun > Grenade by that logic). Some people, though, would probably prefer something that CAN do the most damage (Grenade > Railgun)
Either way, the way I see it: the shotgun is the most consistent and highest average damage weapon...so it should be #1. The pistol, railgun, and rifle all do the same MAX damage, but have different min damages...so I would say Pistol, Rail gun, and Rifle in that order...Which gives us:
 Shotgun
 Pistol
 Railgun
 Rifle
That leaves the Knife and the Grenade and where they should go...while the grenade will do 0 damage 50% of the time, the knife will do the lowest expected value (4.5)...and while the knife CAN do 0 damage some of the time, it won't do 0 damage nearly as often as the grenade. Soooo...if I had to choose, I'd put the Grenade then Knife...so my final order would be:
 Shotgun
 Pistol
 Railgun
 Rifle
 Grenade
 Knife
But again, that's my personal preference based on my definition of "best"...and depending on your playing style, risk level, and other unknown factors (does grenade hit multiple targets??), it may vary.
 1

...a lot...
Spoiler131883 is the count I get.

@Cygnet I think something is wrong with your solution, as one of the requirements is that "18 of us did not drink water."
by your answer, you have 8 that drank nothing, 7 that drank only wine, and 11 that drank beer and wine...which totals 26 people that did not drink water.

Spoiler
1, 2, 3, 4, 8, 9

Spoiler
Alright, this time, I'm getting the smallest number of you who drank all three is 10 (the largest is 13).
Basically it still turns out that there are 0 people that didn't drink anything, so that didn't impact my original answer...but by maximizing the number of people that drank beer and water, I found a different answer for people that drank all three. Here's a breakdown of all 8 groups for reference:
 NOTHING: 0
 WINE ONLY: 3
 BEER ONLY: 0
 WATER ONLY: 9
 WINE/BEER: 15
 WINE/WATER: 11
 BEER/WATER: 6
 ALL: 10
Most of my above approach still worked, except there was a "+ N" (nondrinkers) added to a couple of my equations, which led to a few additional checks by trial and error, but it wasn't too bad.
Hopefully that matches your answer...or I should maybe just stick to my day job...

1 hour ago, bonanova said:
@Pickett It may not impact your solution, but the OP did not intend to say that all of us drank something.
"Class of drinkers" was not meant to preclude anyone from drinking nothing.
Meaning there are eight "classes" of drinkers.Sorry for that. I have this thing about insisting that zero is a number, rather than a denial.
Also, I get a different answer. I'll check my analysis against yours to see why.
I'm guessing it's because I missed this piece of the assumption: "who drank beer and water but not wine were as many as possible".
I'll give it another go with the nondrinkers included and that piece and see what I come up with.

Approach
SpoilerLet's start by identifying the 7 different groups of people involved and label them
 Total people that had only Wine = W
 Total people that had only Beer = B
 Total people that had only Water = H
 Total people that had Wine and Beer = WB
 Total people that had Wine and Water = WH
 Total people that had Beer and Water = BH
 Total people that had Wine, Beer, and Water = WBH
Now, given that, we can write our system of equations based on the problem statement:
EQUATION 1 (total cost of all drinks):
5*(WINE) + 2*(BEER) + 1*(WATER) = 293
5*(W+WB+WH+WBH) + 2*(B+WB+BH+WBH) + (H+WH+BH+WBH) = 293
Simplifies to:
5W + 7WB + 6WH + 8WBH + 2B + 3BH + H = 293EQUATION 2 (total glasses used):
W + B + H + 2WB + 2WH + 2BH + 3WBH = 106EQUATION 3 (total nonwater drinkers):
W + B + WB = 18EQUATION 4 (total wine drinkers):
W + WB + WH + WBH = 39EQUATION 5 (total nonalcohol drinkers):
H = 9EQUATION 6 (ending assumption that wineonly was half the number of beer/water drinkers):
BH = 2WSo, now that we have our 6 equations, we can start doing our algebra:
Plug equation 5 into equation 1 and 2 to simplify slightly:
1) 5W + 7WB + 6WH + 8WBH + 2B + 3BH = 284
2) W + 2WB + 2WH + 3WBH + B + 2BH = 97Plug equation 6 into equations 1 and 2:
1) 11W + 2B + 7WB + 6WH + 8WBH = 284
2) 5W + B + 2WB + 2WH + 3WBH = 97Reorder equation 1 and 2:
1) 2(W+B) + 9W + 7WB + 6WH + 8WBH = 284
2) (W+B) + 4W + 2WB + 2WH + 3WBH = 97Plug equation 3 (W+B = 18WB) into equations 1 and 2:
1) 9W + 5WB + 6WH + 8WBH = 248
2) 4W + WB + 2WH + 3WBH = 79Plug equation 4 (W=39WBWHWBH) into equation 1 and 2:
1) 4WB + 3WH + WBH = 103
2) 3WB + 2WH + WBH = 77Solve for one of the remaining variables (WB):
WB = (103  3WH  WBH) / 4Plug into equation 2:
WH = 1 + WBHAlright, so we now have 1 equation with 2 unknowns, where one of the unknowns is what we're trying to determine! So, we can list out all of the possible values of WBH and WH...which happen to be:
WBH = 1, WH = 2
WBH = 2, WH = 3
...
WBH = 18, WH = 19Anything above WBH = 18 will cause equation 4 to not work...so we can stop there with those 18 possibilities. Now it's just a matter of substituting in each of those values into our equations to find the lowest one that works. I'll spare the gory details, but WBH < 11 results in negative values for other variables at some point...so let's just show trying WBH = 11:
WBH = 11, WH = 12:
Substitute those values into our 4 equations and simplify:
 5W + 2B + 7WB + 3BH = 124
 W + B + 2WB + 2BH = 40
 W + B + WB = 18
 W + WB + = 16
Now substitute 4 into 3 to solve for B:
B = 2...so we know B = 2 and W = 16WB
Plug those two into equations 1 and 2 and simplify
 2WB + 3BH = 40
 WB + 2BH = 22
Solve (WB = 222BH)
You find BH = 4...so given BH = 4, WBH = 11, WH = 12, B = 2, and H = 9, we can find the other two values of WB = 14 and W = 2...at which point we can verify all of the original equations hold true. so therefore, WBH (total number of people who drank all three drinks) is at least 11 given the assumptions we stated. 
Yes...I can.
SpoilerThe smallest number of you who drank all three is 11.
I'll write up the details and approach I took in a little bit.

FWIW
SpoilerThere are only 83 total...and they are all under 100,000,000...
SpoilerAlso going with the "tree" approach that Molly Mae mentioned:
2
23
233
2333
23333
23339
2339
23399
233993
2339933
23399339
239
2393
2399
23993
239933
2399333
29
293
2939
29399
293999
2939999
29399999
3
31
311
3119
31193
313
3137
31379
317
37
373
3733
37337
373379
3733799
37337999
37339
373393
3739
37397
379
3793
3797
5
53
59
593
5939
59393
593933
5939333
59393339
59399
593993
599
7
71
719
7193
71933
719333
73
733
7331
7333
73331
739
7393
73939
739391
7393913
73939133
739393
7393931
7393933
739397
739399
79
797 
Nevermind...misunderstood the OP (again).

Spoiler
AL: Since he removes the lowest value each time, he ends up with all of the values from N+1 to 2*N in his bag. So the total can be calculated as (3N^{2} + N) / 2
BERT: He's removing an even each time...and since he only gets one even every round, he's left with only the ODD numbers left. So the total can be calculated as simply N^{2}
CHARLIE: He's obviously a little bit trickier...I will spare you the ugly details of how I arrived at this, but basically you are calculating the expected value by multiplying the number by the probability that that number will be present at midnight. This calculation can be ultimately simplified down to a sum: ∑ _{x=1 > N} (4x^{2}x) / (N+1)
So...Here's a quick example of their payouts for increasing values of N:
N  AL  BERT  CHARLIE (expected)
1  2  1  1.5
4  26  16  22
10  155  100  135
250  93,875  62,500  83,375
1000  1,500,500  1,000,000  1,333,500
75998  8,663,582,005  5,775,696,004  7,700,940,671.895
250250  93,937,718,875  62,625,062,500  83,500,125,041.563
5000000  37,500,002,500,000  25,000,000,000,000  33,333,334,166,666.794
10000000  150,000,005,000,000  100,000,000,000,000  133,333,335,000,000 
8 hours ago, ThunderCloud said:
I think...
Let a diagonal road connect each of the four towns to the endpoints of a vertical road that runs through the center of the square. Call the length of the vertical road x. Then the total length of all of the roads, geometrically, is
T = x + 2√(x^{2} 2x +2)
Then, dT/dx = 1 + (2x 2) / √(x^{2} 2x +2), which has a zero at x = 1  (1/√3). The total road length, by the above formula, is then ~2.73205.
Special thanks to Cygnet for suggesting there might be a minimum hidden in such a configuration.
I assumed my answer was along the right lines, but I was too lazy to actually find the true minimum with it...thanks for doing the dirty work!

Spoiler
381654729
 3/1 = 3
 38/2 = 14
 381/3 = 127
 3816/4 = 954
 38165/5 = 7633
 381654/6 = 63609
 3816547/7 = 545221
 38165472/8 = 4770684
 381654729/9 = 42406081

How about
SpoilerTotal length: ^{(2*sqrt(5) + 1)} / _{2 }= 2.7361
It would look something like this:
\__/
/ \ 
My brain isn't exactly firing on all cylinders today, but...
SpoilerSo there are infinite number of triangles available, divided into 3 categories (or sets), each with an infinite number of elements
 An infinite number of right triangles
 An infinite number of acute triangles
 An infinite number of obtuse triangles
As a quick example of that, imagine a right triangle with points at (0,0), (0,1), and (1, 0)...you can just move the point on the yaxis up and down however you want to stretch the triangle infinitely.
So then, it depends on how you define "what fraction".

If you look at it from a statistical perspective and ask "if you draw 3 completely random, noncolinear points in the xy plane, what are the odds of that triangle being obtuse?" I'd have to say it's 1/2 for the following reason:
 While there are an infinite number of right triangles, it is statistically a zero percent chance of the 3 random points creating one. Which leaves 2 possible outcomes, each with equal probability.

However, if you are asking for total number of obtuse triangles divided by total number of triangles, I'd say it's indeterminate.
 It's similar to asking "what fraction of points are inside of a circle on a plane?"
 Technically, since there are an infinite number of points inside the circle AND outside of the circle, you could map every single point on the inside of a circle to some point outside of the circle and vice versa...however, conceptually, it's obvious that there should be more points outside of the circle...but that defies the idea that there's an infinite number inside...
As I said, my brain is pretty scattered today, and I'm assuming that's not the answer you were looking for...

Spoiler
 43
 63
 135
 175
 518
 598
 1306
 1676 < Note: v and x have the same value...I'm not sure if that's allowed by the OP.
 2427 < Note y and @ have the same value...I'm not sure if that's allowed by the OP.
BONUS: Not finding a solution...
abcdefgcc = a^{4} + b^{3} + c^{8} + d^{5} + e^{7} + f^{9} + g^{0} + c^{8} + c^{8} OR
abcdefgcc = a^{4} + b^{3} + 3c^{8} + d^{5} + e^{7} + f^{9} + 1
Even with allowing duplicate values for the variables and allowing leading zeros in the number on the left, I'm not finding a solution...

I got
Spoiler57 distinct color patterns. Still double checking my work, though...

How about
SpoilerSeven? something like this: (pardon my poor paint skills):

Spoiler
 He took note of the time on his clock when he left. We'll call this value T_{S '}
 He walked x amount of time to his friends house.
 He took note of the time there. We'll call this value T_{F '}
 As he leaves he notes the time on his friend's clock. We'll call this value T_{F '' }

He knows what time he arrived and what time he left, so he knows he spent y amount of time at his friend's house:
y = (T_{F '' } T_{F '}) 
He then walks x amount of time back home and looks at his clock. This is T_{S ''}. He knows this value would be:
T_{S '' = }T_{S '} + y + 2x 
So with that information, he can determine x (The amount of time to walk home from his friend's house)
x = (((T_{S''  }T_{S}_{'})  (T_{F'' } T_{F'})) / 2) 
He knows the CORRECT current time would be:
T_{F '' }+ x 
Therefore, he has all the info he needs to set the correct time:
T_{F'' }+ (((T_{S''  }T_{S}_{'})  (T_{F'' } T_{F'})) / 2)
QUICK EXAMPLE:
 Let's say his clock showed 12:00 when he left (T_{S '} = 12:00)
 His friend's clock showed the actual current time of 17:00 when he arrived. (T_{F '} = 17:00)
 His friend's clock showed the actual current time of 20:00 when he left. (T_{F ''} = 20:00) therefore y = 3 hours
 His clock shows 17:00 when he arrives home (T_{S ''} = 17:00)

He knows he should set his clock to:
 20:00 + (((17:00  12:00)  3 hours) / 2)
 20:00 + ((5 hours  3 hours) / 2)
 20:00 + (2 hours / 2)
 21:00

I've not heard this one...but I'll take a stab:
SpoilerLet X be the amount Bill originally bought the scooter for.
Bill's total profit (or loss) = $110  X
Example: If he originally bought it for $100, then:
 Bill buys for $100 (total profit = $100)
 Bill sells to Tom for $100 (total profit = $0)
 Tom sells to Bill for $80 (total profit = $80)
 Bill sells to Hermon for $90 (total profit = $10)
So unless we know what he originally bought the scooter for, we can't really calculate his total profit.

I guess I took this question to be something different. Rather than just "what's the smallest positive value you can think of?" for which I would go with something like:
1^{(3↑↑↑↑...↑↑3)}
or in other words: 1 over Graham's number (or simply "G")...but why stop there? why not go with
1/G↑↑↑↑↑↑↑...↑↑↑↑↑↑↑↑GThat interpretation of the OP isn't as interesting or fun because it just gets ridiculous... The way I took the original question was pick the smallest positive INTEGER that no one else picks...what is your strategy when up against 9 others doing the same (all allowed to pick however many tokens as they want)?
So, if that's the case, it becomes a much more challenging problem. Obviously it doesn't make sense to purchase more than 9 tokens, as with 10 the BEST you could hope for is breaking even...I would probably purchase the following numbers (and my rationale next to them):
 1  I'm assuming I will lose this $10, but maybe everyone else will think someone else picks this and therefore no one else does!
 2  Might as well try this one as well, but again, most likely will lose this $10 as well...
 7...ish  Some relatively low number that hopefully no one else picks...I would assume I'd lose this.
 10  Let's assume everyone buys their 9 tokens and picks 19...10 would be the first number that no one would pick...
 16...ish  Again, some pretty low number, that's pretty much just a shot in the dark and you hope no one else picks it.
 42  If by some crazy chance everyone buys 9 tokens and all are duplicated, this is the lowest possible number that wouldn't be duplicated...plus isn't it really the answer to everything?
At this point, I've already bet $60 in hopes of winning $100...and really, the odds are still not in my favor...so to me, it's not worth the risk and so my ultimate strategy would be to not play and "break even" :c)

Spoiler
679: 679 > 378 > 168 > 48 > 32 > 6
Then for what it's worth, here are the minimums with persistence 69 (notice how they all build on each other):
 6788: 6788 > 2688 > 768 > 336 > 54 > 20 > 0
 68889: 68889 > 27648 > 2688 > 768 > 336 > 54 > 20 > 0
 2677889: 2677889 > 338688 > 27648 > 2688 > 768 > 336 > 54 > 20 > 0
 26888999: 26888999 > 4478976 > 338688 > 27648 > 2688 > 768 > 336 > 54 > 20 > 0

Spoiler
Agree with 681...Next up would be:
 September: 9919
 October: 71015
 November: 81114
 December: 8124
SpoilerThe "code" is simply:
 {NUMBER OF LETTERS IN NAME}
 {ORDINAL NUMBER OF MONTH IN YEAR}
 {ORDINAL NUMBER OF FIRST LETTER IN ALPHABET}
For example, for "January", it is:
 7 letters long
 First month in the year (1)
 begins with "J" which is the 10th letter of the alphabet
So January = 7 1 10
Going with that, August would be 6 8 1
Take a guess
in New Logic/Math Puzzles
Posted
Haha, that's kind of fun to wrap your head around...
But really, it should be fairly simple, I think:
Each option has a 25% chance of being randomly picked. That is the correct answer: 25%.
BUT there are two options where "25%" is the answer. This means that the answer that you pick has a 50% chance of being "25%", which is STILL the correct answer regardless of the probability of picking an answer that HAS that value. So the probability that you are correct is 50% which is answer 2...