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Jul-16-06
 | | Sneaky: There is a hallway with 1000 doors, each clearly numbered, and they are all closed. You decide to walk down the hallway exactly 1000 times, opening and closing doors according to a formula. Here's the formula. The first time you walk through the hall, you change the state of every door (if it's closed, you open it -- if it's open, you close it.) The second pass, you only change the state of doors divisible by 2; i.e., you change doors 2, 4, 6, 8, etc. The third pass, you only change doors divisible by 3. Then doors divisible by 4, then 5, then 6, and so forth, until on the 1000th pass you only change the state of doors divisible by 1000 (i.e., the very last door). When you're all done, which doors are open? |
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Jul-16-06
| | Sneaky: <142857x5=714285> That's very interesting. Of course this number looks familiar to you, right? 1/7 = 0.142857 repeating. |
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Jul-16-06
 | | OhioChessFan: LOL, <LoFarkas> it IS that simple. I knew it would be too easy for Chessgames users, but decided to post it anyway. Try it out as a bar bet sometime. You'd be amazed how hard some people make it. |
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| Jul-16-06 | | themadhair: <Sneaky> Nice. Lets see now - any numbers state is determined by the number of factors it has. So, for example, 12 will remain closed due to having an EVEN number of factors (6 factors - 12,6,4,3,2,1). Clearly a door will only be open if it has an ODD number of factors. Consider any arbitrary number X. For every factor F of X we have the factor (X÷F). So X will have an odd number of factors if and only if there exists an F such that F=(X÷F) - in other words if X is the square of some number. Since 32²=1024>1000 and 31²=961<1000 then we can only have 31 square numbers and hence 31 doors that remain open. |
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Jul-26-06
| | OhioChessFan: How about a few anagrams? All of these are famous comics.
1. Dear Darling
2. Babe Thrown
3. Servant Time
4. Limbo Relent
5. Muddier Hype |
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Jul-26-06
 | | Phony Benoni: Comic strips or comedians? |
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Jul-26-06
| | OhioChessFan: Comedians, comediennes. TV/Movies |
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Jul-28-06
| | OhioChessFan: Answer to Stumper of July 15:
A number has 6 different digits. If the last digit is moved to the front, then a new number is formed which is exactly 5 times the old number. What is the original number?Write down the digits in the first number as ABCDEF. The new number is hence to be written as FABCDE. We define G to be ABCDE.
So: 100000F + G = 5 (10G + F)
99995F = 49G
Division by 7 gives 14285F = 7G
Because F is a number with 1 digit, the solutions is G = 14285 and F = 7. We started with 142857. |
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Jul-28-06
| | OhioChessFan: <themadhair> Playing around with the 314 problem in my mind, it seems that if we were talking whole numbers, we'd want to go with (17+19) (16+20), etc. to maximize as closely as possible the squares that add up to 314. That's a little less than the square of 18. I am at a loss, though, to see how there can be a finite answer with the provision they need not be whole numbers. Should we use 17.1 and 19.9?
17.11 and 19.89? At some point, there's an infinite regression toward the square. |
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Jul-29-06
| | OhioChessFan: While you are leaving from your previous ordeal, you notice that another convict is being taken to another jail cell. As you watch the guard and the convict go to the cell, they start talking.
"You do realize your rights, right?" says the guard.
The convict replies, "The judge said something strange, but I didn't understand it. What was it?"
"You are free to go anytime you like," declares the guard, "as long as you fulfill the required conditions. In your case, your door is secured with a ten-digit lock. If you can guess the right number, you are free to go."
"That's easy," says the prisoner. " I can just keep guessing numbers until I guess the right one."
"Even so," says the guard, "it would take you a hundred years to find the right number at the rate of one per second. Of course, you can always look for the hints we give you." He then points at you and proceeds to tell the new prisoner about your imprisonment.
"In addition to the normal amenities, you have a desk and a scientific calculator. Good luck." The guard walks away. You stand there for a few minutes before you realize that you can go. As you turn to leave, you notice that the new prisoner is walking up behind you.
"Wow, this place is great!" he says. "I only had to input one number!" What number did he try?
Note: There is some room for assumption in the question, so different people may in fact come up with different, though similar, answers. |
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Jul-29-06
| | OhioChessFan: ** Answer to July 25 Anagram Stumper **
***
***
1. Dear Darling = Gilda Radner
2. Babe Thrown = Bob Newhart
3. Servant Time = Steve Martin
4. Limbo Relent = Milton Berle
5. Muddier Hype = Eddie Murphy |
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| Jul-30-06 | | zarra: <OhioChessFan> <What number did he try?> Did he try 3153600000? |
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| Jul-30-06 | | TIMER: <themadhair> I don't have a calculator on me right now, but I can give the method to solve the 314 problem: Clearly all the numbers (which are positive) must be atleast 1 (otherwise the multiple decreases) so have at most 314 numbers. Now all the numbers should be the same (to maximise the product) so just find whole number n between 1 and 314 such that (314/n)^n is maximum. n (314/n)^n
1 314
2 24649
etc. |
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Jul-30-06
| | Sneaky: <themadhair> That's the right answer and the best way to think about it. Only perfect squares have the slightly unusual property of containing an odd number of factors. |
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Jul-30-06
| | Sneaky: A monk decides to climb to the top of a tall mountain. There is a single path from the foot of the mountain to the top, but luckily it's a rather pleasant one. It takes the monk exactly 24 hours to get up to the top of the mountain. He leaves at noon on one day and arrives at the top of the mountain at noon the next day. Mind you, the monk does not travel at a constant rate. Sometimes he's walking sometimes he's running, sometimes he stops for a while, eats some fruit, picks some flowers, maybe takes a nap, and so forth, but in the end he averages 1 day for the entire trip. Once he's on the top of the mountain he hangs out there for a week or so, meditating, and then makes his travel down the mountain. He leaves at noon again, and arrives at the foot of the mountain exactly 24 hours later. Again, when he descends the mountain, he moves at a very staggered rate, taking breaks, picking berries, and so forth. NOW THEN ... what do you think of the following hypothesis? "At some point in the monk's travel down the mountain, between his starting and his ending point, he will be on the exact same point in the path at the exact same time of day as he was when he first ascended the mountain." What do you think of my hypothesis? Now I understand, he will be at the starting and ending points at exactly noon, but that's not what I'm talking about (that's why I said BETWEEN his starting and ending points). What I mean is, there might be some creek 2/3rds the way up the mountain, and he gets to that spot at exactly 11:15pm on the way up, and he happens to be right there at 11:15pm on the way down as well. I say there MUST be such a location. I have a lovely proof in mind, do you know what it is? |
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Jul-30-06
| | OhioChessFan: <Zarra> There are a few different numbers a person might come up with, but you have clearly figured out the gist of the problem. |
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| Jul-30-06 | | TIMER: <themadhair> I would go for something like n=116, as 314/116 =157/58 is as close as you can get to e = 2.718... |
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| Jul-30-06 | | ganstaman: <Sneaky> I actually think I saw this before on this site, though it didn't involve a monk. Superimpose the monk climbing the mountain and the monk descending the mountain. So we have the monk climbing and descending at the same time (maybe just play the videos on the same screen). At some point, the 'two' monks must cross paths since they are traveling the same path but in opposite directions. When they do intersect, it obviously has to be at the same time for each since we started their clocks at the same time. In fact, it doesn't even matter if he takes the same time to go up and down. As long as he doesn't finish one trip before the other begins, your hypothesis holds true. |
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Jul-30-06
| | OhioChessFan: What do you sit on, sleep in, and brush your teeth with? |
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| Jul-30-06 | | suenteus po 147: <OhioChessFan> A chair, a bed, and a toothbrush. |
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| Jul-30-06 | | Tomlinsky: <OhioChessFan> My backside, a coma and vigour? |
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Jul-30-06
| | OhioChessFan: <suenteus po 147> and <Tomlinsky> I happened to ask that at church today and nobody got it. I think a few may have called down fire and brimstone on me when I told the answer. |
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Aug-01-06
| | Sneaky: <ganstaman> You have hit the nail on the head. It is interesting to try to concoct strategies where you, as a monk, would make the situation not be true. You can hypothesize "near light speed" for some parts of your journey, try to spend 90% of your time in a single location, etc., but no matter how hard you try you always end up "meeting your ghost" on the trail. I found that delightful, for some reason that I can't explain. If others found it simpleminded and dull, I can understand that position as well. |
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Aug-01-06
| | OhioChessFan: ** Answer to July 29 Stumper **
What number did the prisoner enter? Here's one possible answer. Leap years, beginning choice of number, extra seconds in some days, simple rounding off, all enter into it. Of course, the gist was the prisoner recognizing the importance of the words "Even so, it would take you a hundred years to find the right number at the rate of one per second." There are 86,400 seconds in a day.
There are 365 days in an ordinary year and 366 in a leap year. Leap years ordinarily come every fourth year, except that, in the 100 years starting now, the year 2100 will not be leap. Therefore, starting from now, there are 86,400 * 36,524 seconds in 100 years, or 3155673600. |
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Aug-17-06
| | OhioChessFan: What is the next number in this sequence:
45, 44, 36, 32, 31, 26, 22, 21, ?
A few hints in the next post. |
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