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Posts posted by bonanova


Just to be clear, n = 3 4 5 7 8 9 11 12 13 ..., and CW and CCW alternate untethered (my new favorite word) to parity?

6 hours ago, kaspar said:Rocdocmac is close. You get one day for the 5 minutes, one day for crossing the international date line, and one day for a leap year.
So if you are east of the international date line at 11:59 on the 28th of February, then cross the line and wait a minute or so, it becomes the 2nd of March. On a subsequent leap year these dates could correspond to Sunday and Wednesday.
Nice solve. And the "few years back" is actually two years ago, when Feb 28 was Sunday and March 2 was Wednesday.

The professor writes a problem on the whiteboard, thus:
25  55 + (85 + 65) = ?
He then inexplicably states that, even though you might disagree, the correct answer is actually 5!Explanation?

A few years back while visiting friends, we celebrated the birthdays of their identical twin daughters Joan and Jane, born just 5 minutes apart. Joan had her party on Sunday, and Jane had hers on Wednesday.
Explanation?

This was my take.
SpoilerBert enters a plane with a random empty seat. Chances are 1/100 that it's his, when no one has to move. Else, chances are 1/99 the empty seat belongs to the guy in Bert's seat, so probability that 1 passenger moves is (99/100) (1/99). Surprise!!! That's also 1/100. Chances that two passengers move is (99/100)(98/99)(1/98) = OMG also 1/100. And so on.
What does that mean? There are 100 equally likely cases (1/100 for each) that { 0 1 2 3 4 ... 48 49, 50 51 ... 97 98 99 } passengers have to move. The 99 candidate passengers who move, or not, are indistinguishable (being first to board does not make Al special in any way.) The chances that any passenger will be asked to move varies uniformly from 0/99 to 99/99, with expected value of 49.5/99 = 50%.
Al is one of those passengers.

8 hours ago, plasmid said:When Al gets on the plane, he will do one of three things: (1) he will sit in his own seat, (2) he will sit in Bert's seat, or (3) he will sit in someone else's seat. The odds that he will sit in his own seat (and be guaranteed to not have to move) or in Bert's seat (and be guaranteed to have to move) are equal.
If Al sits in someone else's seat, call that someone else Charlie. Al will have to move if and only if Charlie has to move and claim his assigned seat which Al took. And Charlie will either (1) sit in Al's seat so Al won't have to move, (2) sit in Bert's seat so Al will have to move (note that the odds of picking Al's seat or Bert's seat are again equal), or (3) sit in someone else's seat so that Al and Charlie will both have to move if and only if that someone else has to claim their seat which Charlie took.
This keeps repeating until someone sits in Al's seat (so Al doesn't have to move) or sits in Bert's seat (so Al has to move), and at each step in the process the person who's taking a seat has equal probability of choosing Al's seat or choosing Bert's seat. So the odds that Al has to move are 50/50.
Nice.
One thing I liked about this puzzle is that it's open to clear thinking. Even tho at first it seems too complex.

But actually ...
SpoilerIf Rectangle (a,b) and Circle (r=1) have the same area, then ab = pi.
If they also have the same perimeter, a+b = pi.
Substituting a=pi/b into the second equation leads tob^{2}  pi b + pi = 0
which has no real solutions.
So the premise of the OP is not true.That's because circles maximize area for a given perimeter.
An equalperimeter rectangle must have less area.The most efficient rectangle is a square, with side pi/2 and area pi^{2}/4.
That's (pi/4), or about 78%, of the circle's area.Good one. You had me going for a few days.


Do any of the clues imply that the indicated needles are adjacent?
I would take "on the left side of" to imply that "pain" is adjacent to "seek," while "to the left of" simply means "the other pain" is not to the right of "interesting little creation." But I'd like to be certain of that. Also, is "interesting little creation" one of the needles? If so, you have described nine, not eight, needles.
Thanks.

You have 10 sets of ten coins. One set of the ten is counterfeit, the others are genuine. The genuine coins weigh exactly 0.10 ounces. The counterfeit coins are exactly a 0.01 ounces off, making the entire set of ten coins 0.10 ounces off. You may use an extremely accurate digital scale only once. How do you determine which set is counterfeit?

SpoilerCircle: Area = pi r^{2} Circumference = 2 pi r
Rectangle: Area = ab = pi r^{2} Perimeter = 2 (a + b) = 2 pi r Diagonal = d = sqrt (a^{2} + b^{2})^{1/2} Width = b.2 (a + b) = 2 pi r => (a + b) = pi r => (a + b)^{2} = pi^{2 }r^{2}.
[1] (a^{2} + 2ab + b^{2}) = pi^{2 }r^{2}.
ab = pi r^{2} =>
[2] 2ab = 2 pi r^{2}
[2]  [1] (a^{2} + b^{2}) = pi^{2} r^{2}  2 pi r^{2} = pi r^{2} (pi  2) = ab (pi  2) =>
[3] a^{2}  (pi  2) b a + b^{2} = 0.
a = { (pi  2) b +/ [ (pi  2)^{2} b^{2}  4b^{4} ]^{1/2} } / 2 = f (b)
Ratio of circumference to diagonal = g.
Write g in terms of rectangle width b.
g = 2 pi r / d = 2 (a + b) / ( a^{2} + b^{2} )^{1/2}
g(b) = 2 ( f (b) + b ) / ( f^{2} (b) + b^{2} )^{1/2}

Clue.
SpoilerWhat is the probability that an arbitrary passenger must move? What does that imply?

@ThunderCloud you're homing in on it, but now you're a little high.
@BMAD It's certainly true that if the FIRST child was a boy born on a tuesday, then it's just the prob that the second child is a boy. But ... the OP does not tell you that. That is, "one is a boy" does not imply "my oldest child is a boy." So your "second" child simply means the "other" child.

Not that I know of. Tables of powers, or spreadsheet where different abc values can be simply typed in, or inspired guesswork?
It's not my fav type of puzzle, but some like this type. 

On 4/11/2018 at 1:14 AM, plasmid said:Is there an easier way to reach the solution? If I try to handle the case of an infinite number of princes to choose from, and I deal with the exact formula I was working with initially instead of the simplified approximation that I ended up using to be able to actually calculate an answer, then I get
So this guy Thomas Bruss solved the general stopping problem.
Spoiler(Check out section 2 on page 1388 of the pdf file available in the above link.)
Bruss assigns 1 or 0 to each event in case it's a best case so far. Then the problem reduces to finding the final 1 in the sequence. The probability p_{k} of say the k^{th} event being a 1 is 1/k, making the odds r_{k} = p_{k}/(1p_{k}) = 1/(k1). He then summed the odds starting with the last event and working backward, stopping when that sum first reaches (or exceeds) unity: (Edit): R_{s} = 1/(n1) + 1/(n2) + ... + 1/(s1) >=1. He proved that s is the optimal stopping point. And doing the math he showed that s/n = 1/e and that ((s1)/n)R_{s} (the probability of success) approaches 1/e, as well as n > infinity.
It's really a beautiful result.

2 hours ago, tojo928 said:Think I have the ideal solution here
A (0,0); B (1,0); C(1,1); D (0,1); E (cos(30),sin(30)); F (cos(60),sin(60))
This solution assumes if you are excatly the same distance from points you get to choose. Very slight tweaks to the points can force the longest path
Start at A. B,D,E&F are all 1 mile away so pick E (+1 mile)
At E. B and F are both same distance away (2 sin(15)~.5176) so pick B (+.5176 mile)
At B. A,E,&F are all 1 mile away so pick F (+1 mile)
At F, D is closest (2 sin(15)) (+.5176 mile)
At D, C is closest (+1 mile)
At C, return to A (2^.5~1.414
Total distance walked= 3*(1)+2*(2sin(15))+2^.5~5.449
SpoilerExtremely close to optimal. So close I'm declaring it solved.
SpoilerThe best solution I know of has length of L = (5+√2+√5+√6) / 2 = 5.549+ by moving one of your points (E) to the center of side AB and starting by going from that point to B. Check it out.

I ask people at random if they have two children and also if one is a boy born on a tuesday. After a long search I finally find someone who answers yes. What is the probability that this person has two boys? Assume an equal chance of giving birth to either sex and an equal chance to giving birth on any day.

@plasmid Not sure why derivative failed, but you're right about the result. Nice solve.
SpoilerIt's exact for infinitely many princes, otherwise stop at the floor function of N/e.

SpoilerLet T_{k} = k_{th} triangular number = sum(i=1,k) { i } = k(k+1)/2
y(x) = prod_{(}_{k}_{=1,inf)} { x/2^{k} } = lim_{(}_{k}_{>inf)} { x^{k}/2^{T}_{k} } = lim_{(}_{k}_{>inf)} e^{[}^{k}^{ ln(}^{x}^{)]}/e^{[}^{T}_{k}^{ ln(2)]}
(1) y(x) = lim_{(}_{k}_{>inf)} e^{[}^{k}^{ ln(}^{x}^{)  }^{T}_{k}^{ ln(2)]}
Then if the continued exponent is evaluated left to right,
z(x) = x ^ (x/2) ^ (x/4) ^ (x/8) ^ (x/16) ^ (x/32) ^ .... = x ^ y(x) = e ^ [ y(x) ln(x) ]
(2) z(x) = e ^ [ ln(x) lim_{(}_{k}_{>inf)} { e^{[}^{k}^{ ln(}^{x}^{)  }^{T}_{k}^{ ln(2)]} } ]
For finite x>0, lim_{(}_{k}_{>inf)} { x^{k}/2^{T}_{k} } exists and is zero, since T_{k} dominates k, and y(x) = 0, and z(x) = 1.
Otherwise, from equation (2) z(x) is indeterminate: z(x) is 1 for finite x, but goes with x to infinity for finite k. The problematic term is k ln(x)  T_{k} ln(2). Since there is no satisfactory extension of L'Hopital's rule to multiple variables, we would have to arbitrarily assign a relation between x and k to get a definitive answer. If we let x = k, for example, we still get the above result. But if k increased more slowly, say letting x = ln(k), then z(x) diverges to infinity. Similarly different behaviors occur when x goes to zero.

On 4/2/2018 at 2:17 AM, Molly Mae said:Ah.
The odds of Al moving seem to be 50%. If we test it against a smaller number of passengers, we can see how often these closed loops form and how often Al is in his own assigned seat. It seems as the likelihood of the latter decreases, the likelihood of the former increases appropriately.
Here I have the layout for 3 and 4 passengers. 1 is Al and the last number is Bert (3 for 3 passengers, 4 for 4 passengers). Any time 3 can take the third spot, Al stays put. Any time Al is already in his assigned seat, Al can stay put. Any time the displaced person finds their seat, Al can stay put. Al staying put is denoted with a 'y'. Al moving gets an 'n'.
Al stays put 3/6 times with 3 Passengers:
123  y
132  y
213  y
231  n
312  n
321  nAl stays put 12/24 times with 4 passengers:
1234  y
1243  y
1324  y
1342  y
1423  y
1432  y
2134  y
2143  y
2314  y
2341  n
2413  n
2431  n
3124  y
3142  n
3214  y
3241  n
3412  y
3421  n
4123  n
4132  n
4213  n
4231  n
4312  n
4321  nThese cases are the ones I calculated as well, and led me to the right conclusion. (I had to solve it as it's not mine originally and I did not receive the solution.) It took a little insight to get me thinking along productive lines. See the spoilers in my April 1 post  btw not an April Fools post  to get there.

13 hours ago, DejMar said:My intuitive placement and path of the cabins would be thus:
Let the vertices of the square be A, B, C, D with the diagonals being AC and BD.Let the first cabin [a] be at point A.
Let the second cabin be a walk of 1 mile from A along the diagonal AC.
Let the third cabin [c] be perpendicular to diagonal AC on side CD.
Let the fourth cabin [d] be at or on toward point D at no less distance than point C from cabin ;
this should be approximately 0.4 miles distant from cabin [c].
Let the fifth cabin [d] be at point C.
Let the sixth cabin [e] be at point B.
Finally, a return to the car at cabin [a], point from cabin [e] A requires another walk of 1 mile.
Accumulative walk is approximately 4.8 miles.Not a bad result.
SpoilerThe locations (four corners, side and circular arc) are all correct. But adjusting the last two a bit, and changing the first cabin, in order to alter the sequence, the distance can exceed 5.

12 hours ago, plasmid said:I’m going to make the math easier by
not assuming that there are 100 suitors, but assuming there are a “large” number of suitors. Let’s say it’s the whole kingdom, K, so modifying my previous answer accordingly we have:
P(GCD=1) = N/K
P(GCD=2) = (KN)/K x N/(K1) ~= (KN)/K x N/K = N(KN) / K^{2}
P(GCD=3) = (KN)/K x (KN1)/(K1) x N/(K2) ~= [(KN)/K]^{2} x N/K = N(KN)^{2} / K^{3}
P(GDC=X) ~= N(KN)^{X1} / K^{X}And the odds of picking MWP for any given GCD are still 1/(GCD1) so the odds of picking MWP are approximately
Sum_{[GCD=2 to K]} N(KN)^{GCD1} / (GCD1)K^{GCD}
That doesn’t make it easy enough to just calculate by hand, but it does make it easy enough to make a spreadsheet to handle the calculations. If I make K=100, then this estimate says that the best number of chumps is 37 of the hundred, which would leave (ironically?) about a 37% chance of getting the MWP.
Great job! Want to look at the reciprocal of that number and guess the exact result?

SpoilerThe locus of points equidistant from the origin is a circle, using standard Cartesian distance.
Distance traveled by delta x and delta y (socalled taxicab metric) travel changes the "circle" into a diamond shape of 4 lines, 2 each at 45 degrees and 135 degrees. In the first quadrant it's the line @plainglazed describes, whose centroid (midpoint) is at a (Cartesian) distance of 2 from the origin. That's the answer I originally had in mind  a result that can be obtained just from the knowledge that the series 1 + 1/2 + 1/4 ... converges to 2.
What I ended up asking for was the average (Cartesian) radius of a Taxicab circle. You can get it by integrating the Cartesian distance to the diagonal line as x goes from 0 to 2 sqrt(2) (which one definitely can't do in one's head) or by simulation, as @plainglazed did. It turns out to be a fundamental constant associated with the parabola  the parabola's version of pi, if you will.
Which makes sense, because the distance from the origin to the 45degree line is a parabola if plotted vs angle from 90^{o} to 0^{o}.
The exact value is ln{ 1+sqrt(2) } + sqrt(2) = 2.2955871...
@Molly Mae blazed the trail and @plainglazed nailed it.
Walk the pattern
in New Logic/Math Puzzles
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Assuming the above, we note that
Each ngon bends the path uniquely. For odd n, the bend is given by 180^{o}/n.
For each 4pgon, where p=1 2 3 ..., the bend is 90^{o}/p.
Since the bends are all in the plane, they can be added in arbitrary order.
4pgons (square, octagon, etc., 90/p,  +  +  + ...)
Bend_{4}_{p} = 90^{o}, + 45^{o}, 22.5^{o} , = (90^{o}) Sum_{(}_{i}_{=1, inf)} (1)^{i}^{+1}(1/i) = (90) ln 2 = 62.383246250395077847550890931236^{o}
ngons (n odd, 180/n, + +   + +   ...)
Bend_{n}_{odd} = 180^{o} { 1/3 + 1/5  1/7  1/9 + 1/11 + 1/13  1/15 + ... }
Off the top of my head, I don't know this sum. Here are some thoughts ...
Alternating series are generally easier to deal with. We can sum pairs of terms to do this in two ways. In the first approach, we sum adjacent terms of like sign:
Bend_{n}_{odd} = 180^{o} { (1/3+1/5)  (1/7+1/9) + (1/11+1/13)  (1/15+1/17) + ...}
= 180^{o} ( 8/15  16/63 + 24/143  32/255 + ... (2^{n}^{+1}/2^{2}^{n}1) ...
In the second approach, we sum adjacent odd terms and separately adjacent even terms to obtain two series:
Bend_{n}_{odd1} = 180^{o} (( 1/3  1/7) + (1/11  1/15) + (1/19  1/23) ...)
= 720^{o} ( 1/21 + 1/165 + 1/437 + ... )
And then the even terms pairwise to get
Bend_{n}_{odd2} = 180^{o} ( (1/5  1/9) + (1/13  1/17) + (1/21  1/25) ...)
= 720^{o} ( 1/45 + 1/221 + 1/525 + ... )
Then add the two results.
Final angle that you're facing after doing this forever is { (90)^{o} ln 2 } + Bend_{n}_{odd} (CW is positive)
To be continued ...
I'm leaving now, to plow through some old texts on evaluating alternating series ... ugh.