DejMar
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Spoiler
For Problem #3, clues #4 and #5 are superfluous. One can logically deduce the combination with the first three clues.
For Problem #7, It shall be assumed that wishes that are non-paradoxical could and would be granted by the genie. At the time Frank makes his wish, the possibility of his living for all future time did exist, thus his wish was fulfilled. (If the meaning of words were static, Frank errs in using the auxiliary verb "could" instead of "would" in making his wish). Bruce's wish does not definitively cancel the intent of Frank's wish, as there were both a passage of time between the making of their respective wishes and before the making the first wish of the two were made. The definition assigned to "meant," a past or past participle of "mean," an intent to convey or signify, is logically able to change in meaning as that is what Bruce's wish, in effect does in its granting. Thus Bruce's wish may have no effect whatsoever on Frank's wish. Jack's wish is also expressly past tense. It is possible that both Frank and Bruce were dead sometimes in the past but are currently alive through some miracle or force of magic. It is logical, then, to deduce that without further limitations imposed on and by definitions, there is no definitive answer to the question with "after all is said and done, who of the three is alive and who is dead." Only assumptions can be made, such as reasoning that with the change of definition of "alive", it being "after all said and done" the same as the definition of "dead," and Frank and Bruce are both "alive (synomynous with dead from Bruce's wish)" and "dead." -
It appears what you (rocdocmac) are saying is:
Assume the pizza is a 2-dimensional circle (in a plane of Euclidean space). All cuts are straight lines (i.e., line segments).
The confusion that occurs is with the terms vertical and horizontal in the assumed plane, as they imply all vertical cuts are parallel to the imaginary y-axis and all horizontal cuts, if such were allowed, to be parallel to the imaginary x-axis. Yet, as the cuts are given to be in any direction, there was an implied contradiction to that inference. -
I do believe #3 -- What 3 positive numbers give the same result when multiplied and added together? -- is not grammatically correct.
I believe the question should be -- What three positive numbers give the same result when multiplied together and when added together? The adverb should not be implying each operation occurs together, but should occur for each operation as each operation is intended to be independent, yet gives the same result.
I do agree with Donald Cartmill that #12 also is not written as it should be. -
As problems #9 and #14, I admit to having made errors.
SpoilerThe number of students choosing neither music nor sports for college would be 100 - ((55 - 20) + (44 - 20) + 20) = 21; and, the number of 8's to be painted would be 20.
Problem #12 is one of the problems that can have more than one correct answer. That is, there is more than one way of defining a stroke of a wall clock.
SpoilerA strike might be defined as the length of time from the beginning of a chime to the beginning of the next chime, if one were to follow; and all chimes are of equal length. Under this definition, the six o'clock six strikes given to be 30 seconds would mean each strike was 5 seconds in length. The twelve strikes at midnight would be 60 seconds. Another definition of a strike may define it as the almost infinitesmal point of time where a chime begins. In this latter assumed definition, the problem assigns the 30 seconds from the first chime of the hour through each interval and subsequent chimes to the last chime of that hour. Basically, the length of time is but the measure of the intervals between each chime, and dismissing the interval of time between the hour's final chime to the beginning of the next hour's chime. With this definition, the 30 seconds would be divided by the five intervals and establish that the length of the standard interval would be 6 seconds long. The number of intervals during the midnight series of chimes would be eleven. 6 seconds X 11 intervals = 66 seconds for the length of the midnight's twelve strikes.
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Spoiler
1. Dependant upon what is intended by the ambiguously written question, the value of x could be either 999, where each rational number is computed as n/(n+1) with (n+1) is given to be 1000, or x could be 10000/11 if the fractions are given in a sequence of terms with each nth term having the value n/(n+1).
2. Given each equation as a and b = c, c is (a - b)||(a + b), and it follows that 7 and 3 = 410.
(Note: || is a standard symbol of concatenation)3. For a*b*c = (a+b+c) the three numbers are {1, 2, 3}.
(1*2*3) = (1+2+3) = 6.4. As one of the numbers is 0, the product of those numbers is 0.
5. There is evidence that the problem is not fully transcribed and presented here.
6. 99 + 9/9 = 100
7. The riddle is "why is there no problem enumerated as 7?"
8. If the product is rounded to the nearest whole number, upwards where it is a fraction and equally distant to the two whole numbers it is nearest, then 1/3 of 10 = 3. If the product is always rounded upward to the nearest whole number, then then 1/3 of 10 = 4.
9. 80. [55 - 20] + (100 - 55) = 80 and [44 - 20] + (100 - 44) = 80.
10. Yes. 5! = 125.
12. Assuming that each strike is given at a constant rate, it should take the 1 minute (i.e, 60 seconds) for the clock to strike 12 times to represent the midnight hour. The final stroke of 6 o'clock being at 6:00:30 and the final stroke of midnight being at 00:01:00.
13. 9.
14. There are nineteen 8's represented in the numbers between 1 and 100.
15. Books do not always follow the same numbering system. Given the assumption that the first page of a book is assigned as 1, and using the definition of sum requiring two or more addends, the sum of these two page numbers is 2. The orientation of the book does not change the page number.
16. 600 earrings.
17. Depends upon the units of weight and systems being used. A pound of gold is generally given in troy units and a pound of feathers is generally given in the avoirdupois system. If the units of measure are in the same system, the are of equal weight. If the units correspond to their respective standards, then a avoirdupois pound of feathers weighs more than the troy pound of gold.
18. As the center of the chain is at half the length of the chain, which is also the distance in which the chain dips, the two ends nailed to the wall would be nailed at the same place. The two ends would be zero feet from each other.
19. Little Johnny has a 50% chance of winning with rolling one standard die. His odds are increased to approximately 55.56% with rolling two standard dice. His odds are decreased to approximately 42.13% with rolling three standard dice. The best odds of getting home with any money (100%) is if Little Johnny does not play. If the question is the odds of getting home with "the money", with "the money" being the total of $600, Little Johnny would best gamble using two dice.
20. The problem is not clearly stated. Are the women being transferred as if they are property? Is there a possibility of sharing the number of women gotten or is there a distinct division?
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Spoiler
Yet not a solution due to {b3, d6, g4, e1}.
Appended post to previous.
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Spoiler
It comes from a typo.
Replace {a1, c1, c3, d3} with {a4, b3, g2, h7}.
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Spoiler
Replace {a1, c2, c3, d3} with {a4, b3, g2, h7}.
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A possible solution:
SpoilerRemove the four pieces {a5, c3, d3, c6} and replace them at {a4, e8, g7, g2}.
4 hours ago, rocdocmac said:SpoilerCorrect, no d5. 27 pieces.
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3 hours ago, rocdocmac said:
Looks like there are even more perfect squares on the original layout!
My latest count is 11 squares with 27 pieces being involved and 11 of them found in more than one square.
SpoilerMy own count has also increased, both with squares and pieces. 12 squares and 28 pieces. The four additional squares are: {a6, c3, f5, d8}, {e5, d3, f2, g4}, {b2, c1, d2, c3}, {a5, d3, f6, c8} -- the previously listed being: {a1, c1, c3, a3},
{a1, e1, e5, a5}, {c1, h1, h6, c6}, {a3, d3, d6, a6}, {c3, h3, h8, c8}, {d3, g3, g6, d6}, {c3, d1, f2, e4}, {c3, e1, g3, e5}
The twenty-eight pieces are those at {a1, a3, a5, a6, b2, c1, c3, c6, c8, d1, d2, d3, d5, d6, d8, e1, e4, e5, f2, f5, f6, g3, g4, g6, h1, h3, h6, h8}.
Removing the two pieces at c3 and d3 will leave of these twelve squares but the two {a1, e1, e5, a5} and {c1, h1, h6, c6}. From each of these remaining two one of the four corners of each will also need to be removed. -
16 hours ago, rocdocmac said:Spoiler
I believe the 8 squares you've identified are the following:
{a1, c1, c3, a3},
{a1, e1, e5, a5},
{c1, h1, h6, c6},
{a3, d3, d6, a6},
{c3, h3, h8, c8},
{d3, g3, g6, d6},
{c3, d1, f2, e4},
{c3, e1, g3, e5}
If this is not correct, there are more than 8 squares.
Addendum:
16 hours ago, rocdocmac said:SpoilerThe count of 20 pieces involved is wrong as there are 21 pieces involved in total in corners of these 8 squares:
{a1, c1, d1, e1, h1, f2, a3, c3, d3, h3, g3, e4, a5, e5, a6, c6, d6, g6, h6, c8, h8}.
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The two moves earlier solution:
Spoiler
Let the initial empty peg-hole be #4.
Jumps: {33, 4, 27, 21}
Remaining pegs: ten at {1, 2, 3, 4, 6, 7, 10, 11, 13, 15}. -
On 4/16/2018 at 6:38 PM, bonanova said:
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?
SpoilerThough I find such a puzzle to have some cleverness, the problem is with punctuation. It is missing. The sentence might should end with one or more additional exclamation points, implying emphasis, even though the initial ! character would be the factorial. Of course, the sentence should also be phrased differently. The verb "states" implies that the sentence should end in a full stop (period). This word might should be replaced with "exclaims" to support the additional one or more exclamation points that is suggested.
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Spoiler
One of the solutions (due to symmetry), with the original empty peg-hole at #1 is with the jumps {7, 29, 24, 18, 15, 2}. This pattern of jumps will leave 8 pegs -- 3 in the third and 5 in the 5th (bottom) row of the triangle.
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Spoiler
There is not enough information given to give an answer that can not be logically be uniquely and definitely correct without subject to debate.
"The first (A)merican is the father" does provide the information that the the first American" is male, yet it is not definite whether he is the biological or adopted father. The second American could be either male or female being the biological or adopted father or mother of the son. Other than their relation to the son, the two "parents" own relationship may be, up to the point they have met on the road, strangers -- kinsmen descendents of Noah (if such Biblical belief is accepted) -- to that of brothers, brother-sister, or husband and wife. -
Spoiler
Zero. No illegal moves need be made.
It is possible that the white pieces began at the top. Two pawns on each side advanced to permit each individual piece to move in legal fashion out of their starting places to their mirrored locations, that is other than the King and Queen which would swap their final mirrored positions. No captures needed to be made, and each King can avoid moving into check while it journeys to its "final" locations. Finally, the remaining pawns may be advanced to their "final" positions.
What might be asked is what are the minimum number of legal moves to arrive at the position?
If given that only pawns were moved, all captures were removed, no new pieces were added to the board, and the physical laws were not teased, then the minimum number of illegal moves would be eight. Eight of the pawn could have simply been "hopped" over the pawns into their "final" locations. -
Correction to modified solution. (It appears I incorrectly calculated the second leg's length by failing to divide it by 2, and my approximation of the first leg also needed adjusting.)
SpoilerThe lengths of each leg of the salesman's path should be as follows:
Cabin A to Cabin B = (√(13)/4) miles.
Cabin B to Cabin C = (√(5)/4) miles
Cabin C to Cabin D = 1/2 mile
Cabin D to Cabin E = 1 mile
Cabin E to Cabin F = √(2) miles.
Cabin F to Cabin A = 1 mile.
Total length of the salesman's journey: approximately 5.37 miles. -
Modified solution:
SpoilerLet the following lettered points also be the labels of the six cabins. Let the vertices of the SQUARE be AEDF.
Let the midpoint of DF be C. Let B be the midpoint between CE.
The first leg would be approximately 0.83 miles from cabin A to B.
The second leg would be approximately 1.12 miles from B to C.
The third leg would be 0.50 miles from C to D.
The fourth leg would be 1.00 mile from D to E.
The fifth leg would be approximately 1.41 miles from E to F.
The sixth leg would be 1.00 mile from F to A.
The leg-journey will be approximately 5.83 miles. -
18 hours ago, Thalia said:
How does recognizing divorce contradict engaged men being bachelors? Google shows a bachelor as being a man who isn't and has never been married. So an engaged man would be a bachelor because he's not actually married, a divorced man would not be a bachelor because he used to be married, but as written, it doesn't matter anyway because Green is still married until the suit is settled.
The mistaken inference that one might take is that being engaged does not preclude the possibility that the baseball player had previously been married. That another assumption might be needed to be made in order to solve the puzzle.
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Spoiler
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. -
12 hours ago, Thalia said:
"3. The third baseman lives across the corridor from Jones in the same apartment house."
Normally, I would consider an apartment and a house to be different. But clue number 3 seems to indicate that they are interchangeable.
An apartment and house are different, and clue number 3 does not indicate they are interchangeable, but gives the implication that the house the third basemen and Jones reside contains more than one apartment. Your reference to clue 13 to which it states that Adams lives in a house does not preclude the possibility that Adams may reside in an apartment of an apartment house, not does it state that Adams' sister and Adams lives in the same apartment. From that clue the two siblings could reside in the same or separate apartments -- or in no apartment as the house they reside may, in fact, not be an apartment house, but may be single unit, duplex, or other type of residential house.
An assumption that only a non-bachelor could have children contradicts evidence of reality. Yet, in order to find a solution with the givens, it may be one of the assumptions that is needed to be made. Another assumption that might be taken is that one engaged to be married is a bachelor -- even given that the assumption is partially contradicted by clue 15, with divorce being recognized.
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Spoiler
a※b※c :=: (a×b)||(a×c)||(a×(b+c)-b) :=: (7×5)||(7×2)||(7×(5+2)-5) :=: 35||14||44 :=: 351444
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Two of the many other possible solutions -- the second of which begins with a 12-letter word, and the first having only the beginning or ending letters being erased, thus not forming words with additional white space within the word.
Spoiler9) prelatesS: (archaic) a female prelate or wife of a prelate
8) Prelates: a ecclesiatic of high order
7) relateS: gives an account of; establishes a connection
6) Relate: to give an account of; to establish a connection
5) Elate: to make very happy or proud
4) Late: tardy
3) atE: consumed
2) aT: preposition used to indate a location or position
1) a: indefinite article12) abranchiateS : an organism without branchiae
11) Abranchiate: without branchiae
10) branchiaTe: having branchiae
9) branchiaE: plural of branchia
8) braNchia: a gill or other organ having the same function
7) brachIa: plural of brachium, the part of the arm from shoulder to elbow
6) brachA: (in Judaism) blessing
5) braCh: (archaic) a hound
4) Brah: a brother; a male friend or buddy
3) Rah: (interjection) used as an exclamation of encouragement to a team or player
2) aH: (interjection) an exclamation expressing pleasure, pain, sympathy, etc.
1) a: indefinite article -
Spoiler
ATROPHIES: wasting aways
- A => TROPHIES: awards
- E => TROPHIS: a tropical American tree genus
- S => TROPHI: the mouthparts of an arthropod
- R => TOPHI: a deposit of urates in tissues
- H => TOPI: a pith helmet
- I => TOP: uppermost
- T => OP: operation
- P => O: oh
other possible final sequences: ...>TOI>TO>O; ...> TOI>TI>I'; ...>TOI>OI>I; ...>TOI>OI>O; ...>TOP>TO>O
TOI = the mountain cabbage tree; TO = toward; TI = a musical note; OI = a type of punk music/oy; I = meThere are many possible solutions.
Square in a circle
in New Logic/Math Puzzles
Posted
From my reading of the problem, the question can be re-phrased:
What is the ratio of the sagitta to the radius of a circle where the apothem is equal to the square-root of half the radius squared?
The chosen answer is valid for radius = 1, yet my calculations do not show that to be true for other radius lengths.