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Posts posted by rocdocmac


Spoiler6.74744" x 6.74744" ~ 45.528"^2

Spoiler6.747" (?)

3 hours ago, flamebirde said:Four bodies is correct! "Remains" is used as a noun, so the wording in the OP was not a grammatical error or typo. If the question read "remain" (as a verb), then two left. Or possibly one as CaptainEd wrote. There was no shoot out, TimeSpaceLightForce!

SpoilerI'm afraid not!

There are five people in a room, someone came and killed four. How many remains?

CaptainEd ...
SpoilerSpoton with your most recent octahedral attempt and congrats for solving this one too!
Are you going to try deciphering the dodecahedral case as well? It does take some time if done "by hand", no doubt, but certainly not a too difficult exercise!

SpoilerThe 16 for 3 cycles is correct, but the 20 for 6 cycles needs an update!

SpoilerOne can also explain getting younger by two or three days by time dilation for a body traveling at a speed close to that of light, but this won't do for the time being!

SpoilerA difference of two days (1 in between) is easy to explain ...
At the the time the mother of the twins went into labor, she was traveling by boat, plane or ambulance and crossed (any) time zone in the process . The older twin was was born just before midnight on March 1st of a nonleap year, followed by the second just after midnight (5 minutes later) on 28 February after crossing the time zone. During a leap year the younger twin celebrates her birthday two days before her older twin.
But, could there ever be a timezone difference of more than 24 hours to make one twin celebrate three days after the first (i.e. two days in between according to the OP)? Unless traveling was in space during labor. Otherwise, the twins decided not to celebrate their birthdays on the same day!
SpoilerI am changing " (any) time zone" to the International Date Line since "(any) time zone" won't work!
SpoilerCrossing will be from West to East

35 minutes ago, CaptainEd said:Rocdocmac, we are clearly well matched: I had a misenumeration and a miscount, and I’ll raise you my typographical error.
yes, the 8cycles were duplicated, now I count 12 of them, for a total of 24 out of 3^8.
and, yes, 21658347 is a typo for 21658743, which is a duplicate.
SpoilerCaptainEd got all three right! The probability of no collision for 8 ants starting at cube corners is 24/6561 = 8/2187.
Does anyone wish to try a similar OP with ants on the vertices of a regular octahedron and (pentagonal) dodecahedron?

SpoilerI'm happy with the first set of 12 possibilities (parallel faces) and the number of combinations (6561). Watch out, however, that there are repetitions in the second lot with the crossovers, e.g. 41238567 is essentially the same as 67412385. Also, the entry "21658347" is invalid!
On 4/11/2018 at 11:25 AM, rocdocmac said:Sorry for the "icosagon" in the title! Should read isocagon.
Should have said "isocagon" in the OP is wrong  icosagon in the title needs no change. Double dyslexia!!

Thanks!

Sorry for the "icosagon" in the title! Should read isocagon. 'Twas a dyslexic moment!
16 hours ago, CaptainEd said:SpoilerSo far both correct (a) 1/524288 and (b) 2/27

(a) If an ant is placed on each vertex of a regular isocagon, what is the probability that there will be no encounter between any two ants if all ants start moving simultaneously and randomly (either clockwise or anticlockwise) along the edge to the next vertex without changing direction?
What are the chances of no encounter if there are …
(b) Four ants, each placed on the vertices of a tetrahedron?
(c) Eight ants, starting at the corners of a cube?

On 4/5/2018 at 10:24 PM, BMAD said:I hope there is a nuanced way of solving this besides brute force.
Trust so too!
BN ... is there a way to work it out mathematically? I don't think that logarithms will suffice!

7 hours ago, ThunderCloud said:SpoilerSorry ... nobody said 9! Didn't read carefully!!!
10 minutes ago, rocdocmac said:SpoilerI tend to agree with 8. If the problem is exponential and not linear, how come "9" is right in the middle between 6 and 12? After all, 12*1 = 12, 6*2 = 6, and 8*1.5 = 12, whereas 9*1.5 = 13.5!

SpoilerAnswer:
a = 2, b = 5, c = 9
2^5*9^2 = 2592

Looks like cheerio to this one!

Merci beaucoup Harey et bonne nuit ...

@harey...
Êtesvous Franҫais?

On 3/23/2018 at 12:56 PM, harey said:I do not get what the 12.5 mm are there for. Do you mean after the begin of the first song?
The needle goes down 12.5 mm to the very beginning of the first song, i.e. before the first song actually starts.
SpoilerPay attention to both Bonanova's remarks on 2 February 2018. Otherwise you'll go off track!

Looks like 21 is a bit stiff. Let's change it to 13.


9Balls Cue
in New Logic/Math Puzzles
Posted · Edited by rocdocmac
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A picture ...