phil1882
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Posts posted by phil1882
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this game seems to be impossible for the angel to win. if 50% or more lanterns are dark, the devil flips a dark lantern. otherwise, leave it alone.
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without waiting to hear your answer, I think the answer is
Spoilerthe square root of 1,000,000 = 1000.
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edit:
so my dad was playing with three other people in golf for 9 holes.
The bet is each player puts in $5 dollars. There are 2 games being played. $3 dollars of the $5 goes to who has the most points and $2 of the $5 goes to the trash pot.At he conclusion of the 9 holes my dad was the only winner of the points pot and Larry was the only winner of the trash pot. The points pot should have $12 dollars and the trash pot should have $8.
However my dad didn't have correct change nor did Larry so the pot was left with $10 dollars. Dad took $6 of the $10 and Larry took $4. The next time they played Larry said we did not distribute the money correctly. Larry said Dad should have gotten $7 and Larry $3. Larry's logic is if everyone contributed to the pot, the trash pot should have $8 dollars in it. After subtracting out his contribution of $5, he should have a net winnings of $3 dollars not $4 dollars, and dad $7 dollars instead of 6.
So which way is right???
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so my dad was playing with three other people in golf.
you get 3 dollars for every hole you win, and 2 dollars for odd or unusual shots
each player put in 5 dollars, but my dad didn't have the money nor did one of his fellow golfers.
my dad won all the holes for 12 dollars all together, and the buddy that couldn't pay won all the money from the odd shots.
so because neither he nor his partner contributed to the hole pot, my dad gets 6 dollars from the remaining two players, and his buddy gets 4 dollars for his.
however, anther way to look at is, my dad won 12-5 = 7, and the buddy won 8-5 = 3 for his.
what do you think?
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there is a way however to permute through all possibilities by swapping two letters.
ABC -> ACB -> BCA -> BAC -> CAB -> CBA
ABCD -> ABDC -> ACDB -.> ACBD -> ADBC -> ADCB ->
BDCA -> BDAC -> BADC -> BACD -> BCAD -> BCDA ->
CBDA -> CBAD -> CABD -> CADB -> CDAB -> CDBA->
DCBA -> DCAB -> DBAC -> DBCA -> DACB -> DABC
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i'm not fully sure what you are asking,
yes, no object is perfectly identical to another,
but that doesn't mean they are necessarily worth less/more.
if i have two apples, one shaped like a pear, the other like a cucumber, i still have two apples.
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Spoiler
the pattern mostly seems to be square the number then add 1.
although this doesn't seem to work with 5 to 27.
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your description still doesn't make sense. how can all cuts be both vertical and in any direction and no horizontal. which is it, all vertical (same direction) or any direction?
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i dont really understand the question. your picture seem contrary to your description.
your picture seems to suggest you can cut in any direction, but your description suggests all cuts have to be in the same direction.
i'll solve both.
for lines in any direction: http://mathworld.wolfram.com/CircleDivisionbyLines.html
1/2*(n^2 +n +2), n=500; 125251
5000000 = 1/2*(n^2 +n +2) n=3162
for cuts in the same direction:
each cut adds 1 new region.
for 500 cuts, that 501 rejoins
to get 5000000 peices 4999999 cuts are needed
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i think so.
for the first number n, n= 1, its fairly trivial, as rocdocmac showed. for larger values, the goal is to get it in a ratio of 1, 2 ,3. to do this, first, let's assume they are in some other configuration, with a<b<c. if a = 1, this is also trivial, we can use the binary approach until a,b c is in the right ratio. if a > 1 we can divide b/a and throw out remainder, then use the the binary approach before.
for example, lets say thee are 48 total beans, with 5 on a, 18 on b, and 25 on c. 18/5 = 3, so the 1's digit on b is 1, so transfer from b to a. a = 10 , b = 13, c = 25. now transfer from b to a one more time, a = 20, b = 3, c = 25 C/B, 8, so transfer from a to b, a = 17, b = 6 c = 25, a to b, a = 11, b = 12, c = 25. then form c to b, a = 11, b = 24, c = 13. b is now where we want. than transfer from c to a, a = 22, b = 24 c = 2. then form a to c. a = 20 b = 24 c = 4 a = 16, b = 24 c = 8.
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i'd probably pick
1/tree(3)
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whoops, legitimately forgot about this topic, sorry guys.
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count me in, though i may be rather slow in posting.
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lets do several steps and see if we can develop such a function.
1 H
1 S 1 N
3 H 1 S
2 H 3 S 3 N
9 H 5 S 2 N
12 H 11 S 9 N
31 H 21 S 12 N
54 H 43 S 31 N
so here are our rules.
Hn = 2*Sn-1 +Nn-1
Sn = Hn-1 +Nn-1
Nn = Hn-1
so, combining we have...
Sn = Nn +Nn-1
then
Sn-1 = Nn-1 +Nn-2
thus
Hn = 2*(Hn-2 +Hn-3) +Hn-2
Hn = 3*Hn-2 +2*Hn-3
so...
F(x) = 1 +3x^2 +2x^3 +9x^4 +12x^5 ...
3x^2*F(x) = +3x^2 +9x^4 +6x^5
2x^3*F(x) = +2x^3 +6x^5
F(x)*(1 -3x^2 -2x^3) = 1
F(x) =1/(1 -3x^2 -2x^3)
thus it will be the nth term of this series, whatever that is.
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if i spent a month on it i might be able to do the necessary calculation, don't have that kind of patience though.
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since it been a couple weeks without even a guess, I'll go ahead and post my answer and see if anyone can do better. i get 59.
0 1 0 2 0 3 0 1 0 3 0 2 0 1 0 3 0 1 0 2 0 1 0 4 0 1 0 2 0 3 0 1 0 3 0 2 0 1 0 4 0 1 0 2 0 1 0 3 0 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
0 2 0 1 0 4 0 1 0 ?
51 52 53 54 55 56 57 58 59 60
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heres an alternate version that i came up with,
place the digits 1-n where n is 4 such that no consecutive digit of any step value, starting at step value, repeats.
here's an example where n is 2.
0 1 0 2 1 0 2 1 2 0 1 ?
1 2 3 4 5 6 7 8 9 10 11 12
here starting from 1 and going a step of 1, there are no repeats. starting from 2 and going a step of 2, no repeats, and so on. However there is no way to get 12 without repeating.
your task is to find the max value for 4.
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the first rule of fight club is, you do not talk about fight club.
there are no rules for how rules are made, with the exception that we try to stabilize and civilize society.
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i think you mean...
x: this number is surrounded by no more than x true statements.
what the maximum number of trues you can have, and what would the board look like?
here's my attempt
SpoilerF F F T F
T T F F F
F F T F F
F T F T T
F F F F T
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it seems to me that
T F
F T
should be
1 2
2 1.
if you meant this...
3 2 1 4 1 2 3 3 3 3 3 4 6 4 4 2 3 4 4 3 1 1 2 4 3cannot be solved.
F T start for top left corner
T T
now 1 doesn't work.
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i'd say 6. if you take into account letter frequency, the letters to the left of u far outweigh the letters to the right.
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alright thanks good to know.
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the strings <<>><><>, <><><<>> are considered equivalent, though for the proposes of uniqueness, i put the factors in order from lowest to highest.
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correct.
one of my questions is about addition.
basically my question is, given a number system that uses multiplcation as its basis, (multiplying two numbers in recursive format is just concatenation) an you come up with a method for addition?
ill do 7 as an example
7 is prime.
< 7 >
7 is the 4th prime so replace the 7 with a 4.
< 4 >
4 is composite, its factors are 2 and 2.
< 2 2 >
2 and 2 are prime.
<<2><2>>
the number two is the first prime so cant be reduced further. replace with blank.
<<><>>
Lanterns of Eden.
in New Logic/Math Puzzles
Posted
actually this seems pretty straight forward to me, the angel simply turns off all lit lanterns and wins.