DejMar
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Posts posted by DejMar
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Spoiler
Without changing the arrangement of the letters, I found only four strings of animals:
SUPERNATURAL ANIMALS : honey buzzard
SUPERNATURAL ANIMALS : sea eagle
SUPERNATURAL ANIMALS : Caucasian mountain wild goat
SUPERNATURAL ANIMALS : black tropical American cuckoo
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Spoiler
Quarter is given to be not a portion of the measure or that which is measured, thus it is the measurement, and subsequently, the stone is the material being measured. A quarter is 28 pounds weight, which is approximately 196000 grains. A dram can be anything small, which in this case is the substance of stone. One grain is 1/196000 quarter of stone, and can be said to literally be a grain of sand.
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Spoiler
There are two possible sets of eight words:
PIER, LADS, SENT, TORE, SILL, NAPS, RENT, HOSE
and
LIER, SADS, TENT, SORE, NILL, RAPS, HENT, POSE
PIER - buttress LIER - a person who lies down
LADS - young men SADS - expresses sadness
SENT - dispatched TENT - portable shelter
TORE - rent SORE - tender
SILL - windowsill NILL - not to wish
NAPS - sleeps briefly RAPS - strikes
RENT - torn HENT - sieze
HOSE - flexible pipe POSE - pretend-
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Joyandwarmfuzzies wrote:
"Per the Pythagoren Theorem, 4ac must also be a square, which only occurs when a = c."
4ac may be a square other than when a = c. It is a square when all the factors of 4, a and c can be paired up (4 is is paired up with factors 2*2). As each "a" and "c" are given as odd, neither will have the prime factor 2 as a factor. Yet, "a" may have, for example, the factors 3*3*3 and "c" may simply have the factored 3. This would allow 4ac in this given example to be the square of 18 (=2*3*3).
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"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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n! + 1 is prime for n = {0, 1, 2, 3, 11, 27, 37, 41, 73, 77, 116, 154, 320, 340, 399, 427, 872, 1477, 6380, 26951, 110059, 150209}
Other factorial primes n! + 1 may exist, yet none have been found (as prior to May 2014). From the generally accepted mathematical definition of composite, most other values of the natural numbers n! + 1 are composite. That is, within the domain of natural numbers, bonanova's guess, Aleph-0, is likely correct..
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"Composite" refers to the number having at least two distinct positive integer divisors.
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.
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Roll the d977 and the d709.
977*(d709-1)+d977 [or, alternately, 709*(d977-1)+d709] provides the range of 1 to 692693.
If the result is 692693, re-roll, else divide by 692, and take the ceiling of the result.A re-roll will occur 1/692693 (approximately 0.00014436%) for each roll.
Edit - I see after the fact that Rainman already found this answer.
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Label the faces of the 79-sided die from 1 to 77, with two faces marked re-roll.
Label the 13-sided die from 0k to 12k, where k=77.
Roll the two dice and add the faces, re-rolling the 79-sided die if so indicated.
There is a 2/79, i.e., approximate 2.5316% chance of a re-roll required after each roll, but an approximate 97.4684% chance that the two-dice will result in the random integer required in the first roll. -
@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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Given that non-alphabetic letters are used except for the blank (space) to separate the "words",Given that there are no non-alphabetic characters used, except for the blank (space) used as a word separator,
27x27x27x26 is not the correct solution; nor is 26x25x24x23 as had been selected as the best answer.
The correct count is 26x26x26x26 + 26x26x26 + 26x26 + 26.
a,b,c,d,e,f,g,...,x,y,z = 261 = 26 "words" of 1 letter.
aa,ab,ac,ad,...,zx,zy,zz = 262 = 676 "words" of 2 letters.
aaa,aab,...,zzx,zzy,zzz = 263 = 17,576 "words" of 3 letters.
aaaa,aaab,...,zzzy,zzzz = 264 = 456,976 "words" of 4 letters.
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456,976 + 17,576 + 676 + 26 = 475,254 different "words",thus the magnetic tape does not contain all different "words". At least 24,746 "words" are duplicates
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There is no given restriction on the alphabet/s that is/are used, thus all 500,000 "words" could be different.
The modern English alphabet has 26 letters, yet there are letters of other alphabets that are sometimes used in English, such as ash (Æ), oethel (Œ), and schwau (Ə). In the early modern English alphabet, the ampersand (&) - that is, and, per se, and, was once considered one of the letters of the alphabet, though it was generally used alone, but also possibly in some compound words.
Limited to the 26 letters of the modern English alphabet, absent hyphens and other symbols, there are 475,254 different words possible. For a set of 27 letters, absent other symbols, there are 551,880 different words possible. Hence, the answer to the given problem is subject to a definition of what set of letters/characters/symbols are permitted, and what defines a "letter". -
ERROR...
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Three dice will do the trick.
Take the 7-, 11-, and 13-sided dice.
7*11*13 = 1001, thus, a random integer may be generated uniformly between 1 to 1001, inclusive, using these dice.
Let the faces of the 13-sided die be numbered: 0, 77, 154, 231, 308, 385, 462, 539, 615, 693, 770, 847, and 924.
Let the faces of the 11-sided die be numbered: 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70.
Let the faces of the 7-sided die be numbered: 1, 2, 3, 4, 5, 6, 7.
Roll the three dice, summing the numbers on each of the dice for the random integer. -
Your set of three words could be one of the following:
AEI + DM = (1) AIMED, (2) AMIDE, (3) MEDIA
AEI + DS = (1) AIDES, (2) ASIDE, (3) IDEAS
AEO + DR = (1) OARED, (2) ADORE, (3) OREADAs 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.
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A polynomial of the form (ac)x
2 + (ad + bc)xy + (bd)y2 + (an + mc)x + (bn + md)y + mn can be expressed as having the factors
(ax + by + m) and (cx + dy + n).
Equating the coefficients of the polynomial 2x2 + 3xy - 2y2 - x + 3y - 1 to the expression above:
A: 2 = ac
B: 3 = ad + bc
C: -2 = bd
D: 1 = an + mc
E: 3 = md + bn
F: -1 = mn
WLOG, and seeking only integer coefficients, let m = 1, n = -1
from (E) 3 = d - b, thus, d = b + 3
substituting into ( C),
-2 = b(b + 3)
0 = b2 +3b + 2, which factors to
0 = (b + 2)(b + 1)
b = {-1, -2}
therefore, by again substituting into © and solving
d = {2, 1}
b = -1, d = 2 b = -2, d = 1
3 = a(2) + (-1)c 3 = a(1) + (-2)c
3 = 2a - c 3 = a - 2c
substituting (A), a = 2/c,
3 = 2(2/c) - c 3 = (2/c) - 2c
multiplying both sides by c and factoring
c2 + 3c - 4 = 0 2c2 + 3c - 2 = 0
(c + 4)(c - 1) = 0 (2c - 1)(c + 2) = 0
c = {-4, 1} c = {1/2, -2}
therefore,
2 = a(-4) 2 = a(1) 2 = a(1/2) 2 = a(-2)
a = -1/2 a = 2 a = 4 a = -1
Ignoring the non-integer coefficients, the factors are then
G: (2x - y + 1)(x + 2y - 1), or
H: (-x - 2y + 1)(-2x + y - 1)
(H) is the same as (G) but with each factor multiplied by -1, thus
the factors for the equation can be simply given as
(2x - y + 1), (x + 2y - 1) -
There are two sets of four distinct natural numbers {a,b,c,d} where a*b = c*d = a+b+c+d: {2,15,3,10} and {2,12,4,6}. This is, of course, in addition to the sets of the same numbers ordered differently, i.e., swapping values between the mutliplicands of one expression.or the swapping of the pair with that of the other pair, such as (15, 2, 10, 3).
There is no set of four distinct natural numbers where a*b = c*d = a*d = a+b+c+d, as a*d = c*d would mean that a = c, and a*b = a*d would mean b = d, both of which contradict the given that the natural numbers are distinct. -
An assumption is made that strength and endurance is not altered between contests, no forces
other than a team's combined strength favors either team, and each player puts forward a best
effort to win for their assigned team. An additional assumption is made that the players of
the two mini-teams are the same lost players in which their contest results in a draw.
A+B+C+D+E > F+G+H+I+J (3)
C+D+E = H+I+J (4)
A+B > F+G (4)
A+F < B+G (5)
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F < G -
When taking half the terms of S, the rate at which the infinite series of terms approached 0 was doubled. In addition, the first term of one series of terms, i.e., 1/2, is also set aside before matching up the terms for subtracting one expression from the other, as if the infinite series of the half terms had one more term than the infinite series of the original expression.
Consider the series as finite:
S = 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7.
1/2 S = 1/4 + 1/6 + 1/8 + 1/10 + 1/12 + 1/14
Subtracting 1/2 S from S
1/2 S = 1/2 + 1/3 + 1/5 + 1/7 + (-1/8 -1/10 -1/12 -1/14)
Note, there are subtractive terms that have been failed to be expressed by the problem's falacial logic. In the infinite series where one 1/2 S expressions is subtracted from the other, these subtractive terms will total to the negative equivalence of the sum of the addititive terms, thus the the left-hand side of the expression is not the falacial something, but also is nothing (zero). -
The hypotenuse leg theorem gives that if the hypotenuse of two right triangles are congruent (AX ≡ AX) and a corresponding leg of each are congruent (XB* ≡ XC*), then the triangles are congruent, hence the other legs are also congruent (AB* ≡ AC*). Yet points B and C are not given as equally distant from point A as are the pair of points B* and C*. Furthermore a×b = c×d does not imply |a-b| = |c-d|.
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A flaw is that when the hypotenuse leg theorem is introduced, it equates the (cross-)product of a point and a line with that of another point and line. In the preparatory statements, A, B, or C are defined as points and B* and C* are defined as lines (i.e., line segments). A point is of zero length, thus 0×B* = 0×C*, but the conclusion BB* = CC* does not follow, as the notation BB* and CC* has not been properly defined. If B* were defined as the other endpoint, along with X of a line segment that could be denoted as XB*, then BB* would then could be assumed to be the notation of a line segment. But then BB* would not be either the hypotenuse or leg of any defined right triangle, and thus the introduction of the hypotenuse leg theorem would not follow. A flaw is then using improper, undefined, or inconsistent notation
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"curve" sequence [Fibonacci (2n+1)-1 : n ∈ ℤ
+ ] 1, 4, 12, 33, 88, ...
n = 4
List of points only:
{A}, {B}, {C}, {D}, {A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C,D}, {A,B,C}, {A,B,D}, {A,C,D}, {B,C,D}, {A,B,C,D}
List of line segments only:
{AB}, {AC}, {AD}, {BC}, {BD}, {CD}, {AB,CD}
List of points and line segments mixed:
{AB,C}, {AB,D} {A,BC}, {BC,D}, {A,CD}, {B,CD}, {A,BD}, {AC,D), {A,B,CD}, {A,BC,D}, {AB,C,D} -
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.
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If the distance between a and j are equal to the distances of of the given pair distances being equal, then the chord length (the straight line between two points on a circle) of AB will equal 10*√(2 - 2*cos(3π/5)), which is approximately 16.18034.
The distance between a and j can also be greater or lesser than the distances between the other points, but without any other information, for those two cases the exact distances between A and B is indeterminable. If the distances are rounded between the other pairs of points after determining the distances between the ten points, then the distance between a and j, of course, will not equal the others.
Words of Alphabet
in New Word Riddles
Posted · Edited by DejMar
correct count and word spelling
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