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Posts posted by superprismatic


Does "combined score" mean the total number of chips Moe and Joe have together after 10 rounds?
If so, then the answer is 520, but where is the puzzle?
I'm sure that's not the right answer, so I'm not hiding it in the spoiler.
520 is not the answer because neither Joe nor Moe gets a chip if they turn over cards of the same rank (which, you'll have to admit, will sometimes happen).

I have one question which could make a big difference. It's a bit indelicate, so I'll put it in a spoiler.
Incest. It seems that this may drastically reduce the number of chairs needed.

510
 1
 1

The roots of P are the reciprocals of the roots of Q.

@Bonanova or anyone else who may have an interest in this:
Do you have a proof that that circular arc is the shortest? Conceivably, a piece of a trigonometric curve, or exponential curve, or another conic section may be shorter. The possibilities are endless.

a circular arc, with center at either (1,0) or (0,1), of length .25×sqrt(2)×sqrt(pi) which is approximately 0.62666. The radius is sqrt(2)/sqrt(pi).

I'll bet that the answer is divisible by 7.

Let n be your position in the line, there are n1 people ahead of you. Let A be the event that the people ahead of you all have different birthdays, and let B be the event that you have the same birthday as someone ahead of you. You will win iff both A and B happens. So P(win) = P(A and B) = P(A)*P(BA). We know that P(A) = 365*364*...*(365n+2)/365
^{n}^{1}, and it's easily seen that P(BA) = (n1)/365. So P(win) = 365*364*...*(365n+2)*(n1)/365^{n}. Consider the function f(n) as being your probability of winning in position n. We seek the value for n which maximizes f(n).If we compare f(n+1) to f(n), most factors will fortunately cancel out.
f(n+1)/f(n) = 365*364*...*(365n+2)*(365n+1)*n/365^{n+1}/(365*364*...*(365n+2)*(n1)/365^{n}) = (365n+1)*n/(365(n1)) = (365nn^{2}+n)/(365n365)
As long as this is greater than 1, the function f(n) keeps growing. Solving:
1 < (365nn^{2}+n)/(365n365)
365n365 < 365nn^{2}+n
365 < n^{2}+n
365 > n^{2}n
Since n can only be positive integers, the inequality holds exactly for 0 < n < 20. So f(n+1) > f(n) as long as n < 20, and f(n+1) < f(n) as long as n > 19. Hence n = 20 yields the maximum value for f(n), as hinted by superprismatic's simulation. Entering f(20) into the calculator on my computer yields ~0.0323198575490432954774, which is not far from what the simulation yielded.
Nice analysis! I wish I had thought of that.

gave me the highest probability (0.0323023349) at position 20.

Doug Adams knows.

Here are two such octagons for n=8. I believe there are no others
up to symmetries.
....18..20.... ......10...... 21..19..22..15 ..11......14.. 17..24..23..13 ......09...... ....16..12....
....20..21.... ......09...... 17..19..23..18 ..12......10.. 16..24..22..15 ......13...... ....14..11....
 1

Here's a solution for the maximum
common sum (41) using the integers
from 1 to 16:
....13..07.... ......09...... 04..11..14..12 ..06......03.. 08..16..15..02 ......10...... ....01..05....
I've found 8 distinct solutions (up
to symmetries) for a common sum of 41.
Proof that 41 is maximal:
Let S be the common sum and let C
be the sum of the four closest
numbers to the center of the octagon.
Then, since the numbers making C
are used in three of the sums, whereas
all the other numbers are in only two
of the sums,
8*S  C = 32*n + 16*17
(n comes from the consecutive integers
being n+1, n+2, n+3, ..., n+16).
In this case, n=0. Now, we rearrange
to solve for S to get
S = 4*n + 34 + (C/8)
which means that C must be divisible
by 8. The largest C possible by
adding four different numbers in the
set {1,2,3,...,16} is 56. This makes
S in our case to be 41.

Any t such that 0 ≤ t ≤ e

I wasn't sure what floors+doors+keys means. I took the liberty of interpreting
this as [number of visited floors]+[number of overall keyindoorlock checks]. If this
correct, I realized that the basement floor need not always be checked because
we may have only one key not spoken for by the time we get to consider it. In
this case, we can be sure that that key is the one for the room in the basement.
Similarly, we need only check 9 of the 10 keys in any door. If they don't
unlock it, we can be sure that the key we haven't checked will unlock it.
I used the above reasoning to make two simulations of 1,000,000 trials each.
One simulation, which tried keys in increasing order of how many doors they opened
so far, with ties broken randomly. In this trial, the average score was 385.9.
The second simulation was similar, but tried keys in decreasing order of how many
doors they opened so far. This second one produced an average score of 423.4.
So, my vote goes to my first strategy as the optimal one.

I have 2 questions:
1. Are we to find keys to open all 61 doors? A door need not be locked for us to test a key to it. Or are we
to find keys for only the 60 locked doors?
2. If the basement is technically the eleventh floor, then there must be 10 floors above it with 10 rooms per floor.
That's a total of 100 rooms plus the office in the basement. Is this correct? If so, shouldn't each aboveground
floor have 6 rooms? Otherwise, there would have to be 40 rooms without doors.

Most mathematically minded people know that if P(x) is a polynomial in x
then r is a root of P if and only if xr is a factor of P(x). More generally,
replacing every instance of x^{n} in P(x) by r results in creating the trivial
polynomial 0 if and only if x^{n}r is a factor of P(x).
Since we wish to annihilate that pesky constant term 1 when we replace things in P(x), it's natural to look to roots of 1 thusly:
Let P(x)=x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1. If x were equal to 1, then P(x) becomes
1+11+11+11+1 which is identically 0. This means that x(1), i.e. x+1, is a factor.
Now, P(x)=x·(x^{2})^{3}+(x^{2})^{3}+x·(x^{2})^{2}+(x^{2})^{2}+x·(x^{2})+x^{2}+x+1. So if x^{2} were to equal 1, then P(x)
would simplify to x1+x+1x1+x+1 which is 0. So x^{2}(1), i.e. x^{2}+1, is another factor.
For x^{3}=1, we get P(x)=x·(x^{3})^{2}+(x^{3})^{2}+x^{2}·x^{3}+x·x^{3}+x^{3}+x^{2}+x+1 which simplifies to
x+1x^{2}x1+x^{2}+x+1 or 1x which is not identically 0, so x^{3}+1 is not a factor.
For x^{4}=1, we get P(x)=x^{3}x^{2}x1+x^{3}+x^{2}+x+1 which is identically 0, so x^{4}+1 is another factor.
Since the product of these three factors is of degree 7, we are done.
So, x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1 = (x+1)·(x^{2}+1)·(x^{4}+1)

What kind of time is 00:77:34?
no, that is 00:00:34 the minute zeros lit as LL
Do they always do that? Do the hours also do that sometimes? I think the mechanics of it all should be explained.

12 because the sequence is symmetric when taken modulo any divisor of 12.

Algebra

If there is only one unfaithful wife, the husband would know
on the first night because there is a least one but he knows that
none of the other men's wives are unfaithful.
If there are two, then each man reasons thusly: If my wife is
faithful, then the husband of the unfaithful wife would know that
his wife is unfaithful (using the reasoning of the preceeding
paragraph) on the first night. Since he didn't write her name
on the first night, my wife must be unfaithful. So, on the second
night both men will write their wives name on the board and the
sequence of numbers of names on the board would be 0,2 and no more
names would be added after that.
It's an easy induction argument to show that for N unfaithful
wives the sequence of numbers of names on the board would be
N1 nights of 0 followed by nights of the same N names on the board
thereafter. 
0,0,0,0,0,0,7.

This type of game is widely known as "Penney's Game". The Penney is Walter Penny who made quite a few puzzles in the 60's and 70's.
I posted nearly 50 of his puzzles (from unpublished papers of his) a few years ago on this forum. You can look at some of them if you search
for Penney on this site. They are quite clever puzzles!

If, by the problem you mean that, if we pick subsets in such a way that every subset
is equally likely to be chosen as any other subset, and we label them A and B randomly,
then you have Bonanova's case where a=b=1/2 for the probability that A⊆ B which,
I agree with Bonanova, is (0.75)^{n}.
If, on the other hand, you mean that, if we pick subsets in such a way that every subset
is equally likely to be chosen as any other subset, and we label them A and B and we ask
what is the probability that A⊆ B or B⊆ A (in other words, whether the smaller subset is
a subset of the larger), then the probability is (3^{n}2^{n1})/2^{2n1}.

Putting a single black pearl in one bag and all of the others in the other bag gives you just a shade under 3/4 probability of picking a black pearl
reversed coefficients
in New Logic/Math Puzzles
Posted · Report reply
One possibility you hadn't addressed: What about roots which are 0, and possibly with multiplicity?