DejMar

Right! Lower bound. And to further clairfy for others, the probability that is "virtually certain", as bonanova just stated in the previous posts, applies to using Graham's number as the lower bound. For the set of natural numbers, the probability is so infinitesimally distant from 1, that it is 1.

karthickgururaj, bonanova is correct that the probability is so close to 1 that it is extre-e-e-e-mely improbably that you would pick a number that did not contained the digit 1 from the set of natural numbers where the upper bound was Graham's number. What isn't true is that this probability is 1. Though extremely small, the probability is not infinitesmally small which is a requirement to correctly declare it to be exactly equal to 1. I believe bonanova was just trying to make the point that it is so close to one that you would not have lived long enough, even if you were born with the Big Bang and began your count then, to count the number of 9's past the decimal point in the probability when it finally was a different digit, thus it may as well be treated as 1.

I hope I took the right slice of cake. I agree with karthickgururaj, nice puzzle. (And, thanks to bonanova's puzzle, too.)

The outer regions are areas where the geometric figure is not intersected by another.

(continued from post #31) Yet, serendipitously, the question you posed for Q is not an XOR, but an XNOR, which yielded the desired truth values you expected.There was just a little confusion as a result of calling it an XOR in your solution post. I do want to say, I applaud you for formulating a good question in the solution. And thank you for taking time to explain it to me.

karthickgururaj, There are many articles about logic functions in the internet, almost all agree with the definition of the XOR given in the Wikipedia article "Exclusive or": "Exclusive disjunction or exclusive or is a logical operation that outputs true whenever both inputs differ (one is true, the other is false)." The truth table of the XOR is opposite to the truth table you expected.

karthickgururaj, it did help me find my error. You are correct, it is not an logical conjunction (AND), yet neither is it a logical disjunction (XOR).

The adjective you used for proper, seems improper. A "proper adjective" is an adjective that is derived from a proper noun, such as the adjective Shakespearean comes from the proper noun Shakespeare. An adjective modifies a noun, or describes or assigns an attribute to the noun. "The cake is indescribable", in the example is assigning the attribute that the cake can not be adequately described.

"Indescribable" is defined as meaning "not able to be adequately described". If you say something is "indescribable", you did describe it, only the description was inadequate. The expression is, thus, not a paradox.

As gravity is a force that increases with mass, its force is affected. When an object slows down, its mass inherently increases -- though it may be unnoticeable due to the small scale we usually find such objects so affected. So small that most of our present day instruments will record no increase. Unless the probability is 1, as "more likely" implies that the probability is not, there is a contradiction in the statement. Even with a probability of 1, the possibility can still exist. Take, for example, the probability of picking a random number that contains the digit 1 from the set of natural numbers. This probability is 1, yet it is still possible to pick the number 5 from the set, and the numeral 5 is not composed of a numeral 1. The paradoxical nature of the problem does lie in semantics, and a lack of clear definitions. What do 'unstoppable' and 'unmoveable' mean? As mass and energy are akin as shown in the equation E=mc2 , one can say that unmoveable is when E=0, and unstoppable is when E = ±∞. Both these two "possibilities" do not actually exist with our understanding of the physical nature of the universe, yet mathematically they can each exist, but not at the same time. By definition, if one exists the other cannot. If the question, then is referring to these "possibilities", then it is not a paradox, but a fallacy.

This is incorrect. The truth-table does coincide with an XOR operation, but the compound statement is not formulated as an XOR, but as AND. The boolean result is different for the operands if Q is false and X is left-handed, he/she should raise left hand ([0 0] => 0). If person in chair 1 responds to Q1 by raising his right hand, he is person A, and he is (right-handed or not). OR he is person (B or C) and he is left-handed. If person in chair 2 responds to Q1R Q2 by raising his right hand, he is person B, and he is (right-handed or not) OR he is person (A or C), and he is left-handed. Etc. koren was partly correct in recognizing that the questions do not necessarily identify who the person is. karthickgururaj was on the correct track in recognizing that the solution may rely on self-referential questions.

Thanks for adding the spoiler, bonanova. The reason one was not included was the parenthetical opening statement: I still am not sure what kind of solution is being asked of the BrainDenizens.

To you (and other BrainDenizens), I apologize, gavinksong. Being argumentative (as opposed to being contrary) is almost a required condition of a puzzle maker and solver. Anyone who proposes an answer with systematic reasoning is, by definition, argumentative. I was not trying to be contrary, either, but was seeking definition to the rules, restrictions and guidelines for this particular puzzle; and offered PerhapsCheckItAgain, suggestions on how he may phrase his rules to be more clear and precise. I am not always, precise or correct. But, I do welcome correction. I do become defensive when I am threatened, and, for this, I do not apologize. But I can forgive PerhapsCheckItAgain. I came to Brainden hoping to enjoy solving puzzles, expecting it to be a pleasant place.

Okay. Again, I was incorrect. Partly due to an incomplete definition of the relationship of m and n. For any solution for m,x the same solution would apply for x,n. Thus, for a specific case m = x, the case is for n = m = x. For such, the player identified in the partial solution where m=2 should be correct.

This is true for all m and n > 1?

This is incorrect (for example m=2, n=1). Yes, witzar, I should have added for the condition m=2, such that n > 1.

I do have a valid question, here. Is this not a forum?

I don't have a proof, but I am not so sure that the basic analysis would not apply. Nonetheless,

You are correct, gavinksong. I did make an error. I suppose it is possible to find a function by mapping the the values and using extrapolation to find the polynomial, but I do not know if this has yet been done.

The concatenation operator used in mathematics is symbolized with ||. With the rules mentioning the it was a valid operation, an assumption was made that the symbol was to be used. Of course 66 already uses concatenation, but the operator is not used (visible). This is true of all multi-digit numbers.