DejMar
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Posts posted by DejMar
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Spoiler
Once he is dead, being hung will no longer be something the criminal should be concerned about. How about wishing to live a very, long and happy life and die a natural death.
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One possible scenario:
SpoilerThe ambulance was speeding down the street, as frequently occurs when there is someone in back being urgently transported to the hospital. Jack's car was parked near the patch of black ice that had formed during the night's winter frost. Jack was performing his moonlighting job of driving the ambulance; and, in his approach to the emergency center, hit the ice patch. The ambulance lost traction, and, while Jack attempted to regain control, swerved into the side of his parked car.
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Spoiler
You are either 35 or 36 (depending upon the month and day of your birth).
452 = 2025, being the only near future year that is a perfect square. 36 = 45 - (2025 - 2016).
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Spoiler
As each 6-decimal-digit string can have at most one digit shared positionally between any two strings of digits, there is at most ten 6-decimal-digit strings that can occur for any given position. The answer to the problem is then the same for any decimal-digit string of length n, such that n > 1, which is 102 = 100 strings of n-decimal-digits. The number of different sets of one hundred 6-decimal-digit strings that can be formed is much greater finite number, but does not change the answer to the total number of different ways one can create a combination of six digit numbers, using the decimal digits within the given constraints.
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Spoiler
Before the conversation, there are fifteen possible permutations of the truth-tellers and liars.
Such that T = truth-teller and F = liar, and the order of the set is {Andre, Brian, Cole, Diego}, these are (non-deliniated): {TTTT}, {TTTF}, {TTFT}, {TTFF}, {TFTT}, {TFTF}, {TFFT}, {TFFF}, {FTTT}, {FTTF}, {FTFT}, {FTFF}, {FFTT}, {FFTF}, {FFFT}.From Andre's statement the following can be eliminated as possibilities: {TTTT}, {TTFT}, {TTFF}, {TFTT}, {TFTF}, {TFFT}, {TFFF}; leaving: {TTTF}, {FTTT}, {FTTF}, {FTFT}, {FTFF}, {FFTT}, {FFTF}, {FFFT}.
Then, from Brian's statement the following can be eliminated as possibilities: {TTTF}, {FTTT}, {FTTF}; identifying Andre as a liar and leaving: {FTFT}, {FTFF}, {FFTT}, {FFTF}, {FFFT}.
Subsequently, from Cole's statement the following can be eliminated as possibilities (and Cole is identified as a liar): {FFTT}, {FFTF}; leaving {FFFT}. {FTFT}, {FTFF}.
Finally, from Diego's statement, {FFFT} and {FTFT} are eliminated, identifying Diego as a liar, leaving {FTFF}.
Brian is the only truth-teller, and Andre, Cole, and Diego are liars.
Captain Ed's solution is equally correct, but more concisely stated.
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How would the man even get the gold bar to the bus, if he can only travel (with the gold bar) by bus?
SpoilerA. There is no need of the bus. If the gold bar can be transferred from person to person, as given by "(a) man is give (sic) a gold bar to take to the bank," then he can get the gold bar to the bank by having a queue of people lined from where he receives the gold bar and the bank, thus it can be transferred in the manner of 'passing the water bucket from the well to the fire'. The queue can be made of two people - one other and himself, with each transfer the one who passed of the gold bar walks ahead to a position closer to the bank to receive it, and pass it on.
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@jasen - Rotations and reflections may be considered the same for the visual layout, but not for the assignment of values to specific points. There are twelve different solutions. If one considers rotations and reflections to be pertinent, so one should one allow the eversion of this particular figure due to the placement of the points. In such case there is but one solution.
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Spoiler
1. As the total of (1+2+3+5+6+7)/2 = 12, A+B+C = D+E+F = 12;
and the sum of each circle must be 16.
A B C D E F
1 5 6 3 2 7
1 6 5 2 3 6
2 3 7 5 1 6
2 7 3 1 5 6
3 2 7 6 1 5
3 7 2 1 6 5
5 1 6 7 2 3
5 6 1 2 7 3
6 1 5 7 3 2
6 5 1 3 7 2
7 2 3 6 5 1
7 3 2 5 6 1
2. As the prime factors of the known points are {2, 3, 2*2, 3*3, 2*2*3}, x must be 2*3*3 = 18.
As the square root of the product of 2*3*4*9*12*18 = 216, A*B*C = D*E*F = 216; the product of each circle is 1296 [362]
A B C D E F
2 9 12 4 3 18
2 12 9 3 4 18
3 4 18 9 2 12
3 18 4 2 9 12
4 3 18 12 2 9
4 18 3 2 12 9
9 2 12 18 3 4
9 12 2 3 18 4
12 2 9 18 4 3
12 9 2 4 18 3
18 3 4 12 9 2
18 4 3 9 12 2 -
Spoiler
A B C D E F G H
1 6 8 5 10 4 3 2 -
Spoiler
A B C D E F G H
1 7 8 2 4 6 5 3 -
Another consideration is the possibility that "Only one statement is true" is meant to refer to one statement of each of the pairs of statements. In such a case...
Spoiler...Natalia is in the left cell and the right cell is empty -OR- a lion is in the right cell and the left cell is empty.
In this scenario, with "right" being opposed to "left", the phrase "open the right door and marry Natalia" becomes almost synomynous with "at death's door" and the only safe door to open is the door to the cell on the left with the princess being within it, or it being empty with no princess in either cell -- (unless the lion is a lioness and, as a daughter of the King of Beasts, is considered a princess). -
Spoiler
For the puzzle, it must be assumed the following givens and inferences are fact.
The givens and inferences are:- The Weird King is honorable and true to his word. He makes no false statements.
- The princess is Natalia, the daughter of the Weird King. Natalia is not a lion.
- There are two doors, and behind each is one cell. These two cells are separate areas that share no overlapping space.
- Natalia is in either the left or right cell, or in neither of the two cells. Natalia's occupancy does not extend to more than one cell.
If the first statement, the one on top, "Only one statement is true", was true, then each of the statements written beside the two cell doors would be false.
Enumerating the four statements:1. "Natalia is in this cell" – Natalia is in the cell behind the left door.
2. "The lion is in the other cell" – A lion is the cell behind the right door.
3. "One cell contains Natalia" – Natalia is in one of the two cells.
4. "One cell contains a lion" – A lion is in one of the two cells.
Given that each of the four statements are false, from (1), Natalia is either in the cell behind the right door, or in none of the two cells,
from (2), there is a lion in cell behind the left door, or in none of the two cells; from (3), Natalia is in neither of the two cells; and, from (4), none of the two cells hold a lion. Thus, no cell contains Natalia or the lion. The Weird King has no intention that you die, but would choose to marry his daughter. Choosing the door on the right – as instructed by the given true statement "Find the right door and marry Natalia", would be then be the logical choice if you wanted to marry Natalia.If the first statement on the top was not self-referential, but was factual, then only of of the four statements would be true.
- If (1.) is true, then Natalia is in the cell behind the left door; yet, then (3.), which must be false would be a contradiction by being true -- thus, (1.) is false.
- If (2.) is true, then the lion is in the cell behind the right door; yet, then (4.) which must be false would be a contradiction by being true -- thus, (2.) is false.
- If (3.) is true, then Natalia is in one of the two cells. As (1.) is false, Natalia must be in the cell behind the right door; and, as (4.) must be false, there be no lion in any of the two cells, which would be no contradiction for (2.) being false.
- If (4.) is true, then a lion is in one of the two cells. As (2.) is false, the lion must be in the cell behind the left door, and, as (3.) must be false, would then have Natalia absent from either cell, which would be no contradiction for (1.) being false.
As both (3) and (4) are still possibilities, choosing the door on the right would still result in not being consumed by a lion and would reveal either the princess or no princess being present, but nonetheless, as the Weird King's decision to have his daughter married, the choice of this door would be the only logical solution.
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Spoiler
The uncle brought over 3 additional goats from his own herd. The division of the father's herd was then divided.
1/2 of 48 = 24 goats for the oldest son. 1/4 of 48 = 12 goats for the middle son, 1/6 of 48 = 8 goats to the youngest son, leaving 4 goats. The uncle then took his 3 goats and the gift goat as gratuity for the solution back to his own herd. -
Spoiler
Yes, it is real.
In complex number analysis, exponentiation with respect to the complex numbers is a multifunction. That is, in calculating the value of ii it can be shown to have more than one value. Yet, unlike the inverse of the exponention a real number by an real number, which is also multivalued (e.g., square-root of 4 is {2, -2}), the magnitude of a number raised to a complex value is not always the same. Using de Moivre's theorem, It has shown that the multivalues of ii are equal to e(-π/2 + 2nπ) for any integer n, with the principal value being e-π/2 where n=0. As can be noted, each of these values are real numbers. Using Euler's formula, exi = cos(x) + i*sin(x), it can be shown that one of these multivalues is the real number of approximately 0.20787957635. -
In answer to you second question,
SpoilerI did not assume that the ratio of the hand lengths were 2 and 1 unit lengths . The ratio of the clock hands was stated in the problem. The unit length was simply taken to be the length of the hour hand with the actual length being unimportant. The distance reported is only an approximation for the span between the two tips of the clock hands between those stated times. The actual calculated results were not exactly the same, and I truncated these results to the seven decimal places). The value for which the distance is equal does occur between each of those two minutes, thus there is some symmetry -- yet, I did not bother to discover when the calculated second or sub-second for each minute in which the points were equal, as the smallest unit of time I was concerned with was that of the minute. For each minute, I calculated the Cartesian-coordinates of the tip of the minute hand as travelling 6 degrees along the circumference of the larger circle of radius 2 and the same for the tip of the hour hand travelling 1/2 degree along the circumference of the smaller circle of radius 1 sharing the same radial point. The distance between the two points was then calculated. The "speed" was assumed to be the difference in distance between each minute interval as time was a constant between each of those intervals. The minute intervals with the greatest receding and approaching distances were then assumed to fit the criteria in answering the problem.
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Spoiler
The greatest receding speed occurs at 12:55 when the distance decreases by approximately 0.0959556 units since the preceding minute and the greatest approaching speed occurs at 11:06 when the distance increases by approximately the same amount. The time lapse between the two recordings is 10 hours and 11 minutes.
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On 12/4/2015 at 5:56 AM, BMAD said:
On the first one we had an entire column of false coins. Now in this situation, each column has precisely 8 false coins. What is the fewest amount of weighings needed to find all of the false coins?
If each column has precisely 8 false coins and each column is composed of 8 coins, then...
Spoiler...it would take 0 (zero) weighings, as logic can be used without the scale to know each of the twenty-four coins are false.
..., or, perhaps
On 12/3/2015 at 11:59 AM, BMAD said:There are twenty-four half-dollar coins stacked in three identical piles....
is intended to be interpreted as each pile has twenty-four coins, for a total of 72 coins in the three piles?
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I am guessing I visually transposed one pair of rows in the spreadsheet. 6+6+6+2 does equal 20. Perhaps I was entranced by that twenty-year-old, and took my eyes off the mathematician's sister.
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Request clarification to the definition intended for tangram:
SpoilerPerhaps it is not intended that N = 7, yet the definition of tangram is "a Chinese geometric puzzle consisting of a square cut into seven pieces that can be arranged to make various other shapes."
The value of N aside, there is still a need to clarify what is meant by "that do not share the same (integer units) length of side." Does the qualification mean that each individual length of a side is a unique integer or that each ordered pair of integer sets [(length, width), such that length >= width] is unique?
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Spoiler
Given that the first mathematician was unable to know the ages of the four children, the sum of the ages occurs multiple times for different sets of integer ages where the sum of the reciprocals equals the multiplicative identity. This only occurs where the sum is 22 (which must be the age of the first mathematician's sister). These values are: (8,8,4,2), (6,6,6,2), (12,4,3,3).
Given the oldest two are twins reduces this to two possibilities: (8,8,4,2) and (6,6,6,2). The first set is likely the intended answer, though the second can not be dismissed without consideration. (It is a fact that a child can be born less than a year after an older sibling, thus for a short period they can have the same age). Still, considering that a single solution is sought, the ages themselves are to be considered the compose the twins (as opposed to the children), and thus (8,8,4,2) would be the answer.
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Spoiler
Something may be UP. (UPENDS & UPSTARTS)
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Possibilities:
Spoiler**(2) The disqualified winner could have been the gold medalist, and not necessarily the silver medalist.
(3) The "single event" could have been for two separate games, where the awardee may have been unavailable to have participated in the award ceremony due to injury or illness (or some justifiable reason), and was awarded the earlier games medal during the award ceremony in which he was also awarded one for the "current" game. (Not a likely scenario).
(4) The Olympic event may have been non-sports related, such as for Graphic Works in 1948 in which Alex Digglemann won both the Silver and Bronze (for his design of two separate posters). -
Possiblities:
Spoiler(1) Though a single event, the individual won medals for two separate categories in the event.
(2) The contestant placed third and was awarded the bronze. The silver medalist was eventually disqualified after the initial awarding and the bronze-winner was then awarded the silver. -
Spoiler
Perhaps, MULTIQUAESTIA, a moth genus of the family Tortricidae is a valid answer for LTIQ. Though the word isn't a common English word, the binomial nomenclature of plants and animals is considered international.
Letter/Number Puzzles : BRAIN * DEN = .....
in New Logic/Math Puzzles
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