DejMar

A list of 287, such that each letter appears later in the alphabet to those that precede it in the word: ABET BEGIN CHIMPS DEITY EMPTY GNOW ABHOR BEGINS CHIN DELO ENOW GORSY ABHORS BEGIRT CHINO DELOS ENVY GORY ABLOW BEGO CHINOS DELS ERST HILT ABLY BEGOT CHINS DELT ERUV HIMS ABORT BEGS CHINT DEMO FIKY HINS ABOS BEIN CHINTZ DEMOS FILM HINT ACER BEKNOT CHIP DEMPT FILMS HIPS ACERS BELOW CHIPS DEMY FILMY HIPT ACES BELS CHIRT DENS FILO HIST ACHY BELT CHIRU DENT FILOS HOPS ACKNOW BENS CHIS DENY FILS HORS ADEPT BENT CHIT DEOXY FINO HORST ADIOS BENTY CHIV DERV FINOS HORSY ADIT BEST CHIVY DEWY FINS HOST ADOPT BEVY CHIZ DEXY FIRS IMPS ADOS BIJOU CHOP DHOW FIRST JORS ADRY BIJOUX CHOPS DIMP FIST KNOP AEGILOPS BINS CHOU DIMPS FISTY KNOPS AEGIR BINT CHOUX DIMPSY FLOP KNOT AEGIS BIOPSY CHOW DIMS FLOPS KNOW AERY BIOS CIST DINO FLOR KOPS AGIN BIRSY CITY DINOS FLORS KORS AGIO BIST CLOP DINS FLORY KORU AGIOS BLOT CLOPS DINT FLOW LOPS AGIST BLOW CLOT DIPS FLOX LORY AGLOW BLOWY CLOU DIPT FLUX LOST AGLU BOPS CLOW DIRT FOPS MOPS AGLY BORS CLOY DIRTY FORT MOPSY AHINT BORT COPS DITZ FORTY MOPY AHIS BORTY COPSY DIXY FOXY MORS AHOY BORTZ COPY DOPS GHIS MORT AILS BOXY CORS DOPY GHOST MOST AIMS BRUX CORY DORS GHOSTY MOTU AINS CEIL COST DORT GILPY MOXY AIRS CEILS COSY DORTY GILT NOSY AIRT CELS COWY DORY GIMP NOWY AIRY CELT COXY DOST GIMPS AITU CENS CRUX DOTY GIMPY ALMOST CENT DEFI DOUX GINS ALMS CENTU DEFIS DOXY GIOS ALOW CEPS DEFO DRUXY GIPS ALPS CERT DEFT EGIS GIPSY AMORT CERTY DEFY EGOS GIRT AMPS CHIK DEGS EIKS GIST ANOW CHIKOR DEGU ELMS GLOP ARSY CHIKORS DEHORT ELMY GLOPS ARTY CHIKS DEIL ELOPS GLORY BEFIT CHIMO DEILS EMOS GLOST BEGILT CHIMP DEIST EMPT GLOW

"As soon as there are martians at points A,B,C such that triangle ABC contains the center of the field..." Are the three points, A,B,C any set of three random points on the circumference of the field from an increasing permutation of triangles that would enclose the field's center? That is, if Farmer Brown has not been transported after the appearance of three alien zoologists, a,b,c, but, a minute later, a fourth martian, d, landed, there are four permutations of triangles, such that zero, one or two of the four permutations that involve the four landed martians form triangles that enclose the field's center point. Would each of the triangles that enclosed the field's center be considered triangle ABC (I.e., more than one triangular transportation zone might exist)? Is it possible that a martian may land on an already occupied point [a Dorothy-Gingema event], and are martian's considered point-like beings for calculation of the probability?

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Actually, that's not quite true as 1 is also a positive integer divisor. Composite really just means "not prime". Not quite true. "The number one is a unit; it is neither prime nor composite." Prime and composite are terms that are applied to positive numbers (though the definitions can been extended to include negative numbers -- as associates to the positive, i.e., the negative number is deemed the same prime or composite as the positive number by the extended definition [refer to the Prime Pages FAQ "Can negative numbers be prime?"]). A composite number has been defined as any positive integer that has at least two prime factors.

Edit - I see after the fact that Rainman already found this answer.

@plasmid Ignoring the re-roll part and probability, rolling 2-dice and taking the product does not create a uniform random integer. Some values will occur more often than others in that manner (1x4 & 2x2 both equal 4, e.g.),

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ERROR...

As the game involved was Scrabble, the first set has the highest point value, though the second set has the best chance of hooking for a high score.

Seeing the answer given as (1+sqrt(5))*5, I decided to use the trigonometric identities and a table of exact trigonometric values to express 10·√(2 - 2·cos(3π/5)) without the trigonometric function. As cos(2u) = 2·cos2(u) - 1, and (3π/5) = 2·(3π/10), u can be equated to (3π/10). cos(2·((3π/10)) = 2·cos2(3π/10) – 1, which, as cos(3π/10) = √((5 - √5)/8), the expression can be further be simplified as ¼·(1 - √5). Thus, 10·√(2 - 2·cos(3π/5)) = 5·√(6 + 2·√5) = 5·√(1 + √5)2 = 5·(1 + √5), which is the same as the answer presented in post #5.