[real number and blackboard]

@plasmid, With the givens, I do not believe {2, 2Π, 4 - Π} satisfy the three reals.
 

Substituting your three reals into the equation

a(2) + b(2Π) + c(4 - Π) = 0

...then solving for Π as if it were a variable results in the following equation, which implies that Π would not be trancendental or irrational given a, b, and c are integers:
Π = (2a + 4c)/(1 - 2b)

[Three 8's to make 24 (my modified version)]

(√(√8×√8 + 8))!

(8 - √(8 + 8))!

(-(√(8 + 8) - 8))!

[Three 8's to make 24 (my modified version)]

I wish to clarify two things:

One, when you say "concatenation is not allowed", does that apply to the factorial signs? That is, does the rule eliminate the double factorial?

 
Two, is this rule modification correct?
"An expression isn't valid unless if at least one operation, or set of parentheses, or both

8‼ / (8 + 8) = 24

can be removed, and the expression is still equal to 24."
At a quick initial read I expected it was trying to eliminate solutions that are basically identical due to the communitive property of some mathematical operations, such as X+Y is basically the same as Y+X, and perhaps the associative property, such as X+(Y+Z) = (X+Y)+Z. Yet, a second and more thorough read it basically states that the solution must be enclosed within another solution or an additonal set of parentheses are required. This interpretation seems odd, thus the request for clarification.

[Snake and Ladders]

Let the twenty-seven intersections be labelled as follows:

A( 1,-1,-1), B( 1, 0,-1), C( 1, 1,-1),
D( 0,-1,-1), E( 0, 0,-1), F( 0, 1,-1),
G(-1,-1,-1), H(-1, 0,-1), I(-1, 1,-1),
J( 1,-1, 0), K( 1, 0, 0), L( 1, 1, 0),
M( 0,-1, 0), *( 0, 0, 0), N( 0, 1, 0),
O(-1,-1, 0), P(-1, 0, 0), Q(-1, 1, 0),
R( 1,-1, 1), S( 1, 0, 1), T( 1, 1, 1),
U( 0,-1, 1), V( 0, 0, 1), W( 0, 1, 1),
X(-1,-1, 1), Y(-1, 0, 1), Z(-1, 1, 1)

One solution that allows each nest to be reachable by the snake via a hollow log and each nest to be reachable by 'you' via a ladder is with the following ladder placement:
AB, AJ, CL, DE, EF, E*, GO, IQ, JK, KL, KS, K*, MU, M*, NW, OP, OX, PH, PQ, P*, RS, ST, UV, VY, V*, YZ

Hollow logs remain at the following locations:
AD, BE, BK, CF, DG, DM, EH, FI, FN, GH, HI, JM, JR, LT, MO, NQ, N*, PY, QZ, RU, SV, TW, UX, VW, WZ, XY

As can be observed, each of the twenty-seven intersections are in both the set of ladders and in the set of hollow logs. (Not all ladders are needed, but were added for ease of use).

[6 Squares and a Common Point]

If all six squares lie in the same plane and exactly two common points are shared among any of the five squares, then each of three squares of the six will share one common point with one each of the other three squares.

Labelling these squares, one vertex (corner point) of square-A shares a point on the perimeter with with square-B, one vertex of square-C shares a point on the perimeter with square-D, and one vertex of square-E shares a point on the perimeter with square-F. When selecting any five of these squares only two points will be shared. Labelling and permuting the three shared points one finds that no common point will be shared for all six combinations of five squares for the six squares.

If the squares are not required to lie in the same plane, the both shared points may be anywhere on the perimeter of the squares. And a sixth square may share one or both points with the other five.

[6 Squares and a Common Point]

If all six squares lie in the same plane and exactly two common points are shared among any of the five squares, then each of three squares of the six will share one common point with one each of the other three squares.

Labelling these squares, one vertex (corner point) of square-A shares a point on the perimeter with with square-B, one vertex of square-C shares a point on the perimeter with square-D, and one vertex of square-E shares a point on the perimeter with square-F. When selecting any five of these squares only two points will be shared. Labelling and permuting the three shared points one finds that no common point will be shared for all six combinations of five squares for the six squares.

If the squares are not required to lie in the same plane, the shared points may be anywhere on the perimeter of the squares. And a six square may share one or both points with the other five.

[A bit wrong]

 |  |
 1101- The error bit is located in the first row, second column.
 0011  [(-13+3-12+13)+(-13+4+11-11)+(13-13-2+11)+(-11-3+12+11)] = 0
-1100-
-1101
  ||

[Alice, Bob and Charlie at a fair]

If we assume a uniform distribution (each digit having an equal probability of being displayed), a bell-curve forms as the error of margin increases from 0 to 5, with the number 5 being the digit most likely to be included within the margin of error. (The curve flattens out again as the error of margin increases beyond 5. Yet, the number 5 holds the greatest probability of appearing with a margin of error within a uniform distribution of error of margin.) Thus, for a uniform distribution, the best guess for Alice's posed wager is 5, and for Bob's -- any number is as likely as the next.

For a non-uniform distribution, if we calculate the standard deviation based on the sample set of 17 observations, subtracting 1 from the number of observations to get a more accurate deviation for the unlimited set, we find σ ≈ 3.137, and the nonparametric skew, γ ≈ 0.319. The positive skew indicates the distribution is likely to be skewed to the right. As the mean of the sample space is 5.29, the standard deviation from the mean of the 17 observations is ≈ 0.761. For a non-uniform distribution based on the sample space, the tip of the skewed bell curve lies approximately at 5.761, which is closer to 6.

Given the sample space, a non-uniform distribution appears likely, and due to the deviation and skew (and the fact that 6 appeared thrice in the 17 observations), 6 may be the best guess posed for Alice's generous offered wager. Given the mode is equal to 9, the choice of number to be guessed for Bob's generously offered bet would be 9.

[A bit wrong]

Without any given specifications of what a 4-way binary chip is, I can only envision the question to be asking for a result of symmetry. If this is the case, the error bit is in the second column of the fourth row.

[maximal triangle area]

The sides of a triangle with the maximal area of 1 occurs at

a=1, b=2, c≈2.23606797.

[maximal triangle area]

If the maximul area is 1 for any triangle, then is this equation satisfied: 0<=a<=1<=b<=2<=c<=3? How?

There was no claim that the maximal area is for ANY triangle, but a triangle that satisfies the requirements, i.e., 0<=a<=1<=b<=2<=c<=3, and the inferred restriction that it is in the Euclidean plane, the maximal area is 1. No Euclidean triangle exists where a=1, b=2, and c=3, thus the length of at least 1 side is smaller. By iterating through various values using Heron's formula you can see that the maximal area approaches 1

.

[The 1962-digit number]

The digital root of any number divisible by 9 (other than 0), is 9.

Let a repdigit number be denoted as x(y), such that y is the number of digits and x is the repeated digit.

From 1 ≤ n < 9(11), the digital sum of a number divisible by 9 is < 99. The digital sum of that number is 9.
From 9(11) ≤ n < 9(1(11)), the digital sum of the digital sum of a number divisible by 9 is < 99. The digital sum of the resulting number is 9.

A 1962-digit number is < 9(1(11)), therefore, given n is the 1962-digit number and S(k) is the digit sum function of k, z = S(S(S(n))) = 9.

[Chess Tournament]

With the givens, the result of the round robin chess match was the following:

1st 2nd 3rd 4th 5th 6th 7th 8th
 7   6   5   4   3   2   1   0
As the round robin consisted of 28 matches, the total number of points equals 28. As each player played in no more than 7 games, 7 is the maximum number of points a single player could score. Given that no player had the same number of points as another at the end of the tournament, and given that the player who ended in 2nd place with a number of points that equaled in total to that of the 5th, 6th, 7th and 8th place finishers, no other possibility could result.
The player winning 3rd place won 5 games, having lost to the 1st and 2nd place players, thus he must have won the game against the 7th place player.

[maximal triangle area]

An alternate answer can be given for a triangle in non-Euclidean space. On a curved plane the definition of a triangle can permit each of line segments of the triangle to be of the maximal lengths, i.e. a = 1, b = 2 and c = 3. Yet, it matters infinitismally that these lengths are used without placing bounds on the curvature of the space, as the surface area of the curved space can approach infinity. Simple stated, with non-Euclidean space and no other bounds than the lengths of the line segments the maximal area can be

∞.
I doubt, though, that this is the answer sought, and will assume the problem was meant to be bound in the Euclidean plane, so I shall stick with my answer in post #2.

[maximal triangle area]

For a triangle in Euclidean space, the maximal area is equal to 1.

[Cyberspace village: truth tellers and liars]

SuF SuR MF MR TuF TuR WF WR ThF ThR FF FR SaF SaR

Amy's statement "It is raining and today is Tuesday" is a logical conjunction. It is true only if both its operands "It is raining" and "today is Tuesday" are true, otherwise it is false. Thus, for her statement to be true, it is a rainy Tuesday; and, as Amy tells the truth on rainy Tuesdays this is a valid option. As Amy lies on rainy Mondays, fair Tuesdays, rainy Wednesdays, fair Thursdays, rainy Fridays, and fair Saturdays, these are also valid options. All other conditions are not.

SuF SuR MF MR TuF TuR WF WR ThF ThR FF FR SaF SaR

Bonnie's statement " It is fair or today is Tuesday" can be argued that it is intended as an exclusive disjunction. Due to possible ambiguity, in grammar it is often clarified as such with the conjunctive word 'either'. Yet, an assumption that an exclusive or is meant shall be considered for this solution.

For an exclusive or, the statement is false if either both its operands "It is fair" and "today is Tuesday" are true or if both are false, otherwise it is true. Thus, considering only those days that are yet valid for further consideration, Bonnie's statement is false if it were a rainy day of Monday,  Wednesday, or Friday, or a fair Tuesday.  And Bonnie's statement is true if it were a rainy Tuesday, yet as that is a day she lies, it can be also be eliminated.

SuF SuR MF MR TuF TuR WF WR ThF ThR FF FR SaF SaR

Cybil's statement, like Amy's, is a logical conjunction. It is true only if both its operands "It was not Wednesday yesterday" and "it will not be Wednesday tomorrow" are true. The first operand is true for all days but Thursday, and the second operand is true for all days but Tuesday. The statement is false for the days Tuesday and Thursday. As Cybil tells the truth on fair Thursdays, the day can be eliminated.

SuF SuR MF MR TuF TuR WF WR ThF ThR FF FR SaF SaR

(1) Therefore, it is a Saturday.
(2) The day is fair.

[Cyberspace village: truth tellers and liars]

SuF SuR MF MR TuF TuR WF WR ThF ThR FF FR SaF SaR

Amy's statement "It is raining and today is Tuesday" is a logical conjunction. It is true only if both its operands "It is raining" and "today is Tuesday" are true, otherwise it is false. Thus, for her statement to be true, it is a rainy Tuesday; and, as Amy tells the truth on rainy Tuesdays this is a valid option. As Amy lies on rainy Mondays, fair Tuesdays, rainy Wednesdays, fair Thursdays, rainy Fridays, and fair Saturdays, these are also valid options. All other conditions are not.

SuF SuR MF MR TuF TuR WF WR ThF ThR FF FR SaF SaR

Bonnie's statement " It is fair or today is Tuesday" is a logical disjunction. It is false only if both its operands "It is fair" and "today is Tuesday" are false, otherwise it is true. If it were Tuesday her statement would be true, yet she lies on rainy Tuesdays, thus a rainy Tuesday can be eliminated as a possibility, yet a fair Tuesday would be remain a valid option. If it were a fair Tuesday, a fair Thursday, or a fair Saturday, her statement would also be true. As Bonnie tells the truth on these fair days, these also remain valid options. If it were a rainy Monday, a rainy Wednesday, or a rainy Friday, her statement would be false, and as she tells lies on these rainy days, these remain valid options.

SuF SuR MF MR TuF TuR WF WR ThF ThR FF FR SaF SaR

Cybil's statement, like Amy's, is a logical conjunction. It is true only if both its operands "It was not Wednesday yesterday" and "it will not be Wednesday tomorrow" are true. The first operand is true for all days but Thursday, and the second operand is true for all days but Tuesday. The statement is false for the days Tuesday and Thursday. As Cybil tells the truth on fair Tuesdays and fair Thursdays, it can not be on such a day. As she tells the truth on the rainy days of Monday, Wednesday, and Friday, and on fair Saturdays, these remain valid options.

SuF SuR MF MR TuF TuR WF WR ThF ThR FF FR SaF SaR

(1) Therefore, it is a Monday, Wednesday, Friday or Saturday.
(2) The day is fair if it is Saturday, else it is raining.
 

[Upside Down Cake]

Trying to solve this for an irrational number, I did find that it is possible for the edge of a slice to meet a transition point due to the transposition of the flipped piece. Though it still was not proof, as the example I was looking at did not eliminate the transition point- pair, but only moved it and an additional transition point-pair had been created, and could imagine that as one transition may be eliminated, but multiple transitions may be created. My opinion had changed to that it may be possible for the elimination of all transitions, but I still was searching for a proof whether that was to be always the case. I will check out the video, perhaps the answer is there.

[Covering chessboards with dominoes]

The answer for 3 is "no". Placing a domino so the protruding single squares is covered reveals that an internal square will be unfilled.

The answer to 4 is "no". A 1x4 quadomino will not fill a 10x10 checkered-board. 10 is not evenly divisible by 4.

[real number and blackboard]

(f) all three real values are greater than 0, with two of the coefficients

negative with the third positive of opposite sign to the third, and the sum of the terms with the negative two coefficients of equal sign being equal in absolute value to the third term;

[Covering chessboards with dominoes]

1. Yes.

One configuration is:
.AABBCC5
DDEEFF25
GGHHII26
JJKKLL36
MMNNOO37
PPQQRR47
SSTTUU48
.VVWWXX8

2. Each square may be left over, depending on the arrangement.

3. With an assumption that the squares surrounded by a blue outline are to be left domino-free, and a domino fits inside the border of any two adjacent squares, the answer is "Yes".

4. With an assumption that the squares surrounded by a blue outline are to be left quadomino-free, and a quadomino fits inside the border of a four-square area, the answer is "No".

[real number and blackboard]

It is given that a

₁•r₁ + a₂•r₂ + a₃•r₃ = 0 : r_n ∈ R₀⁺ and a_n ∈ Z with at least one of a_n <> 0. (n ∈ N, an index)

Either:
(a) all three real values equal 0;
(b) two of the three real values equal 0 and the coefficient of the other equals 0;
(c ) one of the three real values equals 0, and the two coefficients of the others equal 0;
(d) one of the three real value equal 0, and the two coefficients of the others are of opposite sign and the terms to which they are part of are of equal absolute value;
(e) all three real values are greater than 0 with one of the coefficients being zero and the signs of the other two
are opposite with the absolute values of their terms being equal; or
(f) all three real values are greater than 0, with two of the coefficients negative with the third positive,
and the sum of the terms with the negative coefficients being equal in absolute value to the third term;

For cases (a), (b), (c ) and (d) at least one 0 already appears on the blackboard, thus only cases (e) and (f) need be examined. For (e) and (f), if an irrational value exists, all terms will be a multiple of that irrational value. With the exception for (e), a distinct irrational number can be assigned the 0 coefficient. As these terms can be scaled by dividing by the common multiple of the irrational value, the solution would be the same as if each were rational.
It has already been noted that by using the Euclidean algorithm, that a finite number of operations of substituting Y – X for Y will result in a value of 0 appearing on the blackboard.
It may only need be proved that the proposition made, that "all terms will be a multiple of that irrational value", with the noted exception, is true.

[Paula and Victor play another guessing game]

A: For even S, there are S/2 numbers in [1,S] range that are greater than S/2, one number that is equal to S/2 and S/2-1 numbers that are less than S/2. If the OP is rephrased so that the first guess is for "not greater than" instead of "less than" then there is exactly 50% chance for the first guess for any even S. In the current phrasing, there a small bias toward x or y to be greater than S/2.

For odd S, the range [1,(S-1)/2] has one less number than range [(S+1)/2,S], so there is a higher chance that x or y is greater than S/2 for odd S.

Conclusion: Victor should always guess that the first number he picks is greater than S/2.

B: Estimate S based on the revealed number (let's call it a₀) and a₁...a_n. The accuracy of the estimate depends on n and there are multiple different ways to do it. One way is to compute the average (a_avg) of a₀...a_n and assume S=2*a_avg. Another way is to find the maximum (a_max) of a₀...a_n use the formula S=a_max+a_max/(n+1)-1. (For other methods to estimate S see http://en.wikipedia.org/wiki/German_tank_problem)

Given the estimate of S and the revealed number, Victor should answer "smaller" if the number revealed is greater that half the estimate of S and "larger" otherwise.

 

There is no specification that the numbers are distinct, yet an implication that they must be. If the numbers do not need be distinct, it is possible for no number to be less and all to be more than s/2. (This can occur if s = 1). For this specific instance, an answer of "more" will always be a 'win' for Victor. In this instance, the hypothesis that Victor should answer "less" if the number revealed is greater that half the estimate of S and "more" otherwise will be correct, as Victor will always answer "more". S=2*a

_avg=2, and the revealed number R=1, R=S/2. But for the given estimate S=a_max+a_max/(n+1)-1, S<=1, and the revealed number R=1, R>=S/2 (yet, considering probability, as n's upper value is unbounded in the positive integers, R>S/2), the hypothesis will be incorrect (though the hypothesis may hold true for most arbitrarily chosen s).

Given the phrasing "Paula then selects n integers....She tells these numbers....", and "...selects two more integers....", there is an implication that s must be greater than 2. By this implication, I believe your analysis, k-man, is generally correct. But is it correct if x and y do not need be distinct? And, other than the case where s = 1, is it correct if s = 2?

 

[Upside Down Cake]

The theory that "there are in fact a finite number of states because the slices do eventually align", can not be true. Such a statement implies that an irrational number is either even or odd, id est, rational.

A rational number can never be irrational.
No amount of rational units of a cycle will result in the alignment of a cut and a flipped cut where the cuts are of irrational measure. And, it is by definition that a number of cuts is a measured in the counting numbers, id est, natural numbers, a subset of the rational numbers.

On the other hand, when expressed in an infinite number of cycles, we are using neither rational nor irrational numbers. Infinity, itself, is not a number. It is used in the calculation of the convergence to some real value, or the divergence toward a unboundless value. After an infinite number of cycles of slices creating sections that become infinitesmally small, each of the top and bottom of the cake will both approach to a state of both being frosted and both being unfrosted. Id est, there must be a divergence toward an unboundless value, as the definition of the state of frosted is the opposite of the state of unfrosted. Thus, even after an infinite number of cycles, there is no solution for an irrational value.

[Upside Down Cake]

QUOTE (from plasmid's 'proof')

D) The cutter reaches the transition point again after N slices and either destroys it with process #1 or leaves it alone with process #2. If the latter, then we start over at point A of this loop and repeat until the transition eventually gets destroyed.
....
Also for completeness: this only proves that all transitions will eventually be eliminated, ...
END QUOTE

plasmid's proof only shows that all transitions will be eventually eliminated. No proof that the number of transitions is finite.
On the contrary, applying the definition of an irrational number, neither the slice points nor the transition points will align (in less than an infinite number of transitions and rotations).

To the question of whether d can include irrational numbers, the answer is yes -- after an infinite number of cut-flip-rotate processes. But, to the answer where d can include irrational numbers after a finite number of cut-flip-rotate process, the answer is no.