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Feb2118
  al wazir: <zborris8: 1. What is the maximum number of squares that a bishop can visit without revisiting any square?> 29. For example, b1a2b3a4b5a6b7c8d7c6d5c4d3c2d1e2
f3e4f5e6f7g8h7g6h5g4h3g2h1.
<2. One clock says 11:45. The other clock says 12:05. What is the correct time if one of the clocks is 9 hours and 55 minutes fast, and the other clock is 2 hours and 25 minutes slow?> 11:45 = 705 minutes or 1,425 minutes after midnight; 12:05 = 5 minutes or 725 minutes after midnight; 9 hours and 55 minutes = 595 minutes; 2 hours and 25 minutes = 145 minutes. Let T denote the correct time on the first clock and T' denote the correct time on the second clock. For which of the following is T ≡ T' (modulo 720)? a) T = 705  595 = 110 ≡ 830; T' = 5 + 145 = 150 ≡ 870. b) T = 705  595 = 110 ≡ 830; T' = 725 + 145 = 870 ≡ 150. c) T = 1,425  595 = 830 ≡ 1,550; T' = 5 + 145 = 150 ≡ 870 . d) T = 1,425  595 = 830 ≡ 1,550; T' = 725 + 145 = 870 ≡ 150. e) T = 5  595 = 590 ≡ 130 ≡ 850; T' = 705 + 145 = 850 ≡ 130. f) T = 5  595 = 590 ≡ 130 ≡ 850; T' =1,425 + 145 = 1,570 ≡ 850 ≡ 130. g) T = 725  595 = 130 ≡ 850; T' = 705 + 145 = 850 ≡ 130. h) T = 725  595 = 130 ≡ 850; T' = 1,425 + 145 = 1,570 ≡ 850 ≡ 130. Answer: e), f), g), and h). So the correct time is 2:10 am or 2:10 pm. <3. I once had the chance to pass out books to a number of children. If I gave each child 3 books, then I would have one book left over. But if I decided to give each child 4 books, then I would have no books left by the time I got to the last child. How many books did I start with, and how many children were there?> Let N be the number of children and B the number of books. Then B = 3N+1 and B = 4N 4. So N= 5 and B = 16. <4. Which of the book titles are real, and which ones are fake?
https://www.boredpanda.com/funnybo... Most of the titles are unfamiliar to me, but I recognized the one by Malcolm Bradbury, so it's real. I therefore conjecture that all are real. <5. What are the common names of Domenikos Theotocopoulus, and FrancoisMarie Arouet?> I don't know the first one but the second is Voltaire. 

Feb2118
  TheFocus: <Domenikos Theotocopoulus> = El Greco 

Feb2118
  Marmot PFL: <<zborris8: 1. What is the maximum number of squares that a bishop can visit without revisiting any square?> 32 I believe. Bd4a1c3e1d2c1b2a3b4a5d8c7b8a7b6c5
d6f8e7h4f6h8g7h6g5e3f2g1h2g3f4e5 

Feb2118
  al wazir: <Marmot PFL: 32 I believe.> Of course! For some inexplicable reason I was restricting my moves to one square at a time. That's why my bishop never visited a8, e8, or f1. But it's easy to include them in the sequence I posted above if I relax that spurious restriction: b1a2b3a4
e8
b5a6b7c8d7c6
a8
d5c4d3c2d1e2f3e4f5e6f7g8h7g6h5g4
h3
f1
g2h1 

Feb2118
  john barleycorn: the phrase "visiting a square" needs to be clarified. Is the starting square a "visited square"? Moving the bishop from d4 to a1 means he has not "visited" c3 and b2? 

Feb2118
  john barleycorn: of course, after that clarified the answer to "what is the maximum number of squares ...?" cannot be an example like <al wazir: <zborris8: 1. What is the maximum number of squares that a bishop can visit without revisiting any square?> 29. For example, b1a2b3a4b5a6b7c8d7c6d5c4d3c2d1e2 f3e4f5e6f7g8h7g6h5g4h3g2h1. > 

Feb2118
  Marmot PFL: I took "visiting a square" to mean that once a square was touched it couldn't be crossed again. Could be interpreted in other ways too. Toughest piece is the knight
https://en.wikipedia.org/wiki/Knigh... 

Feb2118
  john barleycorn: <Marmot PFL: I took "visiting a square" to mean that once a square was touched it couldn't be crossed again. Could be interpreted in other ways too. ...> I took it as moving one square diagonally. The bishop moving from a1 to c3 for example crosses b2 to which he can legally move which in my understanding means moving from a1 to c3 means the bishop "visited" b2. A knight on a1 jumps to b3 or c2 and cannot legally move to any other field. 

Feb2118
  Marmot PFL: < A knight on a1 jumps to b3 or c2 and cannot legally move to any other field.> I have seen that knight's tour given as a timed chess test but I don't think I ever timed it myself. With black pawns on c3, c6, f3, f6 it can get frustrating. Carlen could probably whip right through it.
http://www.chesschat.org/showthread... 

Feb2118
  john barleycorn: < Marmot PFL: I took "visiting a square" to mean that once a square was touched it couldn't be crossed again. ...> According to your understanding would
<a4e8b5a6b7c8d7c6a8d5> be a valid move order? 

Feb2118
  Marmot PFL: No, because c6a8 would cross b7 

Feb2118
  john barleycorn: <Marmot PFL: No, because c6a8 would cross b7> Thank you, also a8d5 would cross b7 and c6. So we have the same understanding. In fact, with this in mind we can already conclude that starting the bishop (let's say the dsb) on a square other than a1 or h8 will give a maximum of less than 31. 

Feb2118
  Marmot PFL: Thought it could get to 32 squares from most squares, except if it starts on b2 you get 31, missing a1 (or h8 if it starts on g7). 

Feb2118
  john barleycorn: Starting at b2 you sure miss a1. but how to get to h8 without either missing to visit f8 or h6? 

Feb2118
  zborris8: I won't comment on the answer to the chess puzzle <#1> as some are still sorting it out. But I will say congratulations to <ughaibu> who solved at least <#3>, and to <al wazir> who solved at least <#2>,<#3>, and <#4> by setting up the problems when necessary. Finally, nice teamwork by <al wazir> and <TheFocus> in finding the names listed in <#5>. Good job to all! There are a number of solutions listed for the chess puzzle. <john barleycorn> requests the definition of "visiting a square". The puzzle's definition: <"A square is 'visited' if the bishop starts from it, lands on it, or moves through it."> This definition was omitted by mistake. To make up for the chess puzzle's omission, here is a <supplemental>: White to move and mate in 3, using each of his three pieces once and only once:
click for larger view 

Feb2218
  al wazir: <zborris8: ... here is a <supplemental>: White to move and mate in 3, using each of his three pieces once and only once.> There is an obvious mate in 2: 1. Rd6 Kc8 2. Ra8#, or 1. Ra8+ Kxa8 2. Rc8#. But the answer you want is 1. Rd6 Kc8 2. Ka7 Kc7 3. Rac6#. 

Feb2318
  zborris8: <al wazir: There is an obvious mate in 2: 1. Rd6 Kc8 2. Ra8#, or 1. Ra8+ Kxa8 2. Rc8#. But the answer you want is 1. Rd6 Kc8 2. Ka7 Kc7 3. Rac6#.> Yup. Good job! The given answer to the Bishop problem was 29 squares: f1 g2 h3 g4 h5 g6 h7 g8 f7 e8 d7 c8 b7 a6 b5 a4 b3 a2 b1 c2 d3 c4 d5 e6 f5 e4 f3 e2 d1. Was 31 accomplished or not?
********
<Numbers Puzzle> What is the longest string of "1"s and "0"s that meets the following conditions: A. No "1" appears exactly two digits after another "1". B. No "0" appears exactly three digits after another "0". 

Feb2418   Nisjesram: <awn: <jessicafischerqueen:
<Nitwit Jam Head> Will you please cease your inane, rambling, profoundly uneducated spamming. It is irritating beyond comprehension. You are and have always been the most addle minded nitwit ever to post at this website. Permanent ignore. Hopefully enough will follow suit so you get auto blocked.> I knew someone was going to say this sooner or later, and it is spot on. <Tga> is not going to be happy that <Nizzle> left his dirty toilet paper all over his forum.> I am leaving chessgames.com forever from today (as regular poster. I may post under special circumstances which I detail below) , because instead of thanks , I get insults. I showed <big pawn> how stupid omv argument was , I also showed him that he did not know teachings of jesus or philosophy, he did not know difference between 'personal god' and Absolute (he did not even know what Absolute is till I mentioned this 'concept' to him) And instead of thanks , I get such insults.
Today is my last day on chessgames.com. as regular poster.Now, I may post here only if someone insults me or thanks me or needs my help in physics/maths/economics etc. Or if I need to inform people that johnlspouge has endorsed my refutation of stupid omv argument. Thank you
Namaste 

Feb2418
  john barleycorn: < Nisjesram: ...
Today is my last day on chessgames.com. as regular poster. ...> Thank you
Namaste 

Feb2518
  al wazir: <zborris8: What is the longest string of "1"s and "0"s that meets the following conditions: A. No "1" appears exactly two digits after another "1". B. No "0" appears exactly three digits after another "0".> Suppose the first digit is 0: 0XXXXXXX...
By B the fourth digit must be 1: 0XX1XXXX
By A the second digit must be 0: 00X1XXXX
By B the fifth digit must be 1: 00X11XXX
By A the third digit must be 0: 00011XXX
But if the sixth digit is 1, that violates A. If it is 0, that violates B. Therefore the string stops at five digits. Suppose the first digit is 1: 1XXXXXXX...
By A the third digit must be 0: 1X0XXXXX
By B the sixth digit must be 1: 1X0XX1XX
Suppose the second digit is 0: 100XX1XX
By B the fifth digit must be 1: 100X11XX
Then the fourth digit must be 0: 100011XX
But if the seventh digit is 1, that violates A. If it is 0, that violates B. Therefore the string stops at six digits. Now suppose the second digit is 1: 110XX1XX
By A the fourth digit must be 0: 1100X1XX
By B the seventh digit must be 1: 1100X11X
By A the fifth digit must be 0: 1100011X
But if the eighth digit is 1, that violates A. If it is 0, that violates B Therefore the string stops at seven digits. The answer is seven. 

Feb2518
  zborris8: <al wazir> That's the one, zero errors. Good job! I enjoyed the setup as much as the solution. 

Feb2518
  WorstPlayerEver: <Nisjesram>
Thanksfor your posts!
Well, I post because this site is great for research. Obviously I have to restrict myself; because I don't feel obliged to do the homework of the GMs, who do very little fruitful homework themselves anyway. So it seems they'd rather write their obese booklets. If I were in charge I would so and so eliminate 96% of the population. But that's another topic. In other words: don't worry! :) 

Feb2518
  al wazir: <zborris8: That's the one, zero errors.> In hindsight, it's obvious that the answer has to be seven. In fact, if you replace "two" by m and "three" by n in the statement of the problem, where m and n are any integers satisfying 1 < m < n, the answer is 2m + n and the longest possible string has the form 1..10..01..1. 

Feb2518
  al wazir: Let's get back to chess.
Without looking at a chessboard (or a diagram of one), answer the following questions: 1. What is the largest number of ♔s that can be placed on the board so that no ♔ is threatening a square on which another ♔ is located? 2. What is the largest number of ♕s that can be placed on the board so that no ♕ is threatening a square on which another ♕ is located? 3. What is the largest number of ♖s that can be placed on the board so that no ♖ is threatening a square on which another ♖ is located? 4. What is the largest number of ♗s that can be placed on the board so that no ♗ is threatening a square on which another ♗ is located? 5. What is the largest number of ♘s that can be placed on the board so that no ♘ is threatening a square on which another ♘ is located? 6. What is the largest number of ♙s that can be placed on the board so that no ♙ is threatening a square on which another ♙ is located? 

Feb2518
  Marmot PFL: With queens I know that it's 8. Someone evidently has proved that the answer is n queens on an nxn board as long as n is 5 or larger. 



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