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DividedQuantum
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Registered: 12/06/13
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PROBLEM #33
Quote:
A traveler finds himself in town without funds; he expects a large check to arrive in a few weeks. His most valuable possession is a gold watch chain of twenty-three links. To pay for a room, he promises to give the landlady as collateral one link a day for twenty-three days. He wants to damage the chain as little as possible. He can give the landlady a separate link each day, or he can give her one link on day 1, then on day 2 take back the link and give her a chain of two links. On day 3 he can give her the single link again and on day 4 take back all the links she has and give her a chain of four links. All that matters is that each day she must have the number of links corresponding to the number of days. What is the smallest number of links the traveler must cut off to meet his agreement for the full twenty-three days?
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night_shift
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Registered: 01/02/15
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Re: mathematical brain teaser thread *DELETED* [Re: DividedQuantum] 1
#22507850 - 11/11/15 02:10 AM (8 years, 2 months ago) |
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Post deleted by night_shift
Reason for deletion: [delete]
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DividedQuantum
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Re: mathematical brain teaser thread [Re: night_shift]
#22508575 - 11/11/15 09:40 AM (8 years, 2 months ago) |
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Quote:
night_shift said:
Quote:
DividedQuantum said: PROBLEM #33
Quote:
A traveler finds himself in town without funds; he expects a large check to arrive in a few weeks. His most valuable possession is a gold watch chain of twenty-three links. To pay for a room, he promises to give the landlady as collateral one link a day for twenty-three days. He wants to damage the chain as little as possible. He can give the landlady a separate link each day, or he can give her one link on day 1, then on day 2 take back the link and give her a chain of two links. On day 3 he can give her the single link again and on day 4 take back all the links she has and give her a chain of four links. All that matters is that each day she must have the number of links corresponding to the number of days. What is the smallest number of links the traveler must cut off to meet his agreement for the full twenty-three days?
Cut off two links: the fourth and the eleventh from the start, thus producing two single links (2 x length 1) and three chains with lengths 3, 6 and 12.
Solved by brute force, I admit. I started with checking 0 links removed, then 1 link removed... silly, I know, as both these cases could not possibly be the solution. But, by writing down all the possible combinations in a certain order, I now have an idea how to approach similar problems without having to use brute force, so yay me.
Correct!
  
Great job!
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DividedQuantum
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Registered: 12/06/13
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PROBLEM #34
Quote:
Mr. Smith is driving at a steady clip on the highway; Mrs. Smith is seated next to him. "Those annoying signs for Flatz beer seem to be regularly spaced along the road. I wonder how far apart they are?" he comments. Mrs. Smith glances at her watch and then announces the number of Flatz beer signs they pass in 1 minute. "What an odd coincidence!" Mr. Smith exclaims. "When you multiply that number by ten, it exactly equals the speed of our car in miles per hour."
Assume the car's speed is constant, the signs are equally spaced, and Mrs. Smith's minute begins and ends with the car midway between two signs. How far is it from one sign to the next?
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extreme


Registered: 04/05/11
Posts: 9,340
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Every 1/6 mile or .16 repeating??
._l__l__l_.
The points are the starting points, the l's are the signs, the underscores are the distance between the signs. The distance from the start to the first sign and end to the last sign combine to equal 1 full unit between signs. I think whatever speed you plug in comes up with the same answer? I tried 30 mph and 60 mph to make the math easy (60 mph = 1 mile/minute). The faster you go, the more signs you see. I just did a picture lol; didn't really go too far when it came to creating the equation.
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DividedQuantum
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Posts: 9,819
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Re: mathematical brain teaser thread [Re: extreme]
#22517591 - 11/13/15 10:58 AM (8 years, 2 months ago) |
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Quote:
extreme said: Every 1/6 mile or .16 repeating??
._l__l__l_.
The points are the starting points, the l's are the signs, the underscores are the distance between the signs. The distance from the start to the first sign and end to the last sign combine to equal 1 full unit between signs. I think whatever speed you plug in comes up with the same answer? I tried 30 mph and 60 mph to make the math easy (60 mph = 1 mile/minute). The faster you go, the more signs you see. I just did a picture lol; didn't really go too far when it came to creating the equation.
Correct!
  
You're right -- the speed is irrelevant. Outstanding!
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DividedQuantum
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PROBLEM 35
Quote:
Bill, a mathematics student, and his friend John, and English major, usually spin a penny on the bar to determine who will pay for each round of beer. One evening Bill says, "Since I won the last three spins, let me give you a break on the next one. You spin two pennies and I'll spin one. If you have more heads than I have, you win. If you don't, I win." "Gee, thanks," replies John.
In previous rounds, when one penny was spun, John's probability of winning was, of course, 1/2. What are his chances under this new arrangement?
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Sun King



Registered: 02/15/14
Posts: 4,069
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t t t t t h t h t t h h h t t h t h h h t h h h
Same odds
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DividedQuantum
Outer Head


Registered: 12/06/13
Posts: 9,819
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Re: mathematical brain teaser thread [Re: Sun King]
#22517857 - 11/13/15 12:02 PM (8 years, 2 months ago) |
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Quote:
Sun King said: t t t t t h t h t t h h h t t h t h h h t h h h
Same odds
Correct!
 
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DividedQuantum
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Registered: 12/06/13
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PROBLEM 36
Quote:
Two missiles speed directly toward each other, one at 9,000 miles per hour and one at 21,000 miles per hour. The missiles start 1,317 miles apart. How far apart are they 1 minute before they collide?
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Sun King



Registered: 02/15/14
Posts: 4,069
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500 miles
30000/60 = 500
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DividedQuantum
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Registered: 12/06/13
Posts: 9,819
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Re: mathematical brain teaser thread [Re: Sun King]
#22517923 - 11/13/15 12:15 PM (8 years, 2 months ago) |
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Quote:
Sun King said: 500 miles
30000/60 = 500
Correct!
 
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DividedQuantum
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PROBLEM 37
Quote:
Here are two multiplication problems that result in the same solution (3,634):
158 x23 ----
79 x46 -----
Note that the digits 1 through 9 are used once.
Can you rearrange the same nine digits and create two other problems that will yield a larger identical solution? Please note that you must keep the same pattern; that is, the first problem must be a three-digit number multiplied by a two-digit number; the second problem, a two-digit number multiplied by a two-digit number.
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extreme


Registered: 04/05/11
Posts: 9,340
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Ahh that one's tough haha. I've already got a lot of scratch paper used up! I think I see most of the patterns that are required but now I'm still just trying to fill them in correctly. I did come up with this...
138 x69
27x54
SCRATCH THAT I THINK I GOT IT WOOHOO!! I was just putting those numbers down wondering why they wouldn't work lol but then I remembered to switch the equations around I still have them in their double up pairs.
138x27 and 69x54 = 3762!
Thanks for this thread I'm starting to not feel so stupid after all
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DividedQuantum
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Re: mathematical brain teaser thread [Re: extreme]
#22521488 - 11/14/15 09:08 AM (8 years, 2 months ago) |
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That is incorrect. You have to come up with at least two solutions that are the same. So four equations, two and two, and the product of each pair is the same number.
Even though your answer is not one of the solutions I have, it appears that it is correct. So you just have to come up with one more. Good job on the one you got.
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extreme


Registered: 04/05/11
Posts: 9,340
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Dang, I must've misread that. I'm not so smart after all 
At this point I suppose I could just start dividing numbers by 3762 until I start getting some hits. That sounds like a job for another night, though 
I guess if my result isn't apart of one of your answers then maybe it doesn't even have another equation I'll give some others a shot to see if they can solve it. My vacation is this week so maybe I can spend it finishing this later haha.
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DividedQuantum
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Re: mathematical brain teaser thread [Re: extreme]
#22528621 - 11/15/15 06:20 PM (8 years, 2 months ago) |
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I'm sorry, I wasn't clear. You have to come up with two sets. So, you got one. It wasn't one of the solutions I have, but I checked it, and it's right. You beat the system.
So you just have to come up with one more pair and you get full credit.
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DividedQuantum
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Registered: 12/06/13
Posts: 9,819
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Quote:
Here are two multiplication problems that result in the same solution (3,634):
158 x23 ----
79 x46 -----
Note that the digits 1 through 9 are used once.
Can you rearrange the same nine digits and create two other problems that will yield a larger identical solution? Please note that you must keep the same pattern; that is, the first problem must be a three-digit number multiplied by a two-digit number; the second problem, a two-digit number multiplied by a two-digit number.
Well I guess it's been long enough, so here are the solutions I have:
174x32 = 5,568 96x58 = 5,568
584x12 = 7,008 96x73 = 7,008
532x14 = 7,448 98x76 = 7,448
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DividedQuantum
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PROBLEM 38
Quote:
In poker, a straight flush -- five consecutive cards of the same suit -- beats four of a kind. In each suit, a straight flush can start with an ace, a deuce, or any other card up to a 10 (the ace may rank high or low), making ten possibilities in all. Since there are four suits, there are four times ten, or forty, different hands that are straight flushes. There are only thirteen different four-of-a-kind hands. So, if there are thirteen four-of-a-kind hands and forty straight flushes, why does a straight flush beat four of a kind?
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DieCommie

Registered: 12/11/03
Posts: 29,258
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It probably has to do with the fact that a straight flush requires 5 cards while four of a kind requires 4. With four of a kind there are thirteen possible combinations of the four cards, but then there are 48 possible fifth cards. That means there are something like 13*48=624 possible four of a kind hands, making it more likely than a straight flush.
Im hung up a bit though, because this means that the statement you posted is wrong. There are thirteen possible combinations of four of a kind, there are 624 possible four of a kind hands...
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