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3 hours ago, 1949threepence said:

As you can see, it didn't work.

Mike,

I just did it with a toothpick, it does work. Try warmer water maybe?  I'd just run some hot water and it was still warm. Shouldn't make a difference though.

IMG_2928.jpg.3698eb46e9b49dde08441b9dbe3264bf.jpgIMG_2929.jpg.2c387167990620d9dfb2194b3a913788.jpgIMG_2930.jpg.04428bc7e0bdfb3d2c5ade57d1b1c59c.jpg

 

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3 minutes ago, 1949threepence said:

For a few seconds I was slightly thrown between the dog trotting and running, but had to assume that the dog would be doing a constant 5mph at all times. But it was always ever about the total distance/time run by the dog, whether in a straight line or back and forth. I calculated as follows - maybe more cumbersome than some.

7:15 to 8:15 1st man covers 2 miles

8:15 to 9:15 1st man covers another 2 miles making 4 miles in total.

8:15 to 9:15 2nd man covers 3 miles

9:15 to 10:15 1st man covers another 2 miles making 6 miles in total

9:15 to 10:15 2nd man covers another 3 miles making 6 miles in total 

Therefore the two men are parallel at 10:15am

Meanwhile the dog started out at 8:15 moving at 5mph. At 8:55am, he catches up with the first man having by that time covered 3.33 (recurring) miles. The first man having also covered 3.33 miles by 8:55am

Between 8:55am and 10:15am the dog continues to move at a constant speed of 5mph. As a result, he covers a further 6.66 (recurring) miles. 

Effective total distance covered by dog = 10 miles 

It's actually even easier than that! You're right that the second man catches up with the first at 10:15. That's exactly two hours after he set out. The dog is running at 5mph, so in 2 hours he runs 10 miles.

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A logician with some time to kill in a small town decided to get a haircut. The town had only two barbers, each with his own shop. The logician glanced into one shop and saw that it was extremely untidy. The barber needed a shave, his clothes were unkempt, his hair was badly cut. The other shop was extremely neat. The barber was freshly shaved and spotlessly dressed, his hair neatly trimmed. Why did the logician return to the first shop for his haircut?

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1 minute ago, Peckris 2 said:

A logician with some time to kill in a small town decided to get a haircut. The town had only two barbers, each with his own shop. The logician glanced into one shop and saw that it was extremely untidy. The barber needed a shave, his clothes were unkempt, his hair was badly cut. The other shop was extremely neat. The barber was freshly shaved and spotlessly dressed, his hair neatly trimmed. Why did the logician return to the first shop for his haircut?

If you want to get something done - ask a busy man.

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2 minutes ago, Peckris 2 said:

It's actually even easier than that! You're right that the second man catches up with the first at 10:15. That's exactly two hours after he set out. The dog is running at 5mph, so in 2 hours he runs 10 miles.

I knew my way was too cumbersome and detailed. 

 

 

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1 minute ago, Rob said:

If you want to get something done - ask a busy man.

That may well be true, but the puzzle doesn't specify which of the two was busier.

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He might think that the barber with the neat hair cut used the other barber since there are only two of them in town.

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8 minutes ago, Peckris 2 said:

A logician with some time to kill in a small town decided to get a haircut. The town had only two barbers, each with his own shop. The logician glanced into one shop and saw that it was extremely untidy. The barber needed a shave, his clothes were unkempt, his hair was badly cut. The other shop was extremely neat. The barber was freshly shaved and spotlessly dressed, his hair neatly trimmed. Why did the logician return to the first shop for his haircut?

One of maybe two possible reasons that I can think of: a) the hair of each barber must have been cut by the other. Therefore the logician favoured the barber with untidy hair as he cut the hair of the neatly trimmed barber. In that case the untidy surroundings, lack of shave and unkempt clothes are red herrings.

The other possible reason, b) is that the logician favoured the untidy premises and unkempt barber, on the basis he must be a lot busier than the other guy, because more popular, giving better haircuts.    

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Just now, 1949threepence said:

One of maybe two possible reasons that I can think of: a) the hair of each barber must have been cut by the other. Therefore the logician favoured the barber with untidy hair as he cut the hair of the neatly trimmed barber. In that case the untidy surroundings, lack of shave and unkempt clothes are red herrings.

The other possible reason, b) is that the logician favoured the untidy premises and unkempt barber, on the basis he must be a lot busier than the other guy, because more popular, giving better haircuts.    

a) is correct - as you say, the other factors (unshaven, untidy, scruffy) are classic misdirections.

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Four members of a band are walking to a night concert. They decide to take a shortcut, but must cross a bridge. Luckily they have one flashlight. Because of the varying size of their instruments, it takes each member a different amount of time to cross the bridge - it takes the first person one minute, the second person two minutes, the third person five minutes and the fourth person ten minutes. They must cross the bridge in pairs, travelling at the slower speed so if the one minute person went with the ten minute person, it would take a total of ten minutes. Since there is only one flashlight, one person must come back across the bridge, then another pair can cross. They only have 17 minutes to cross the bridge and still get to the concert on time. What order should they cross to get everyone across and get to the concert?

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A=1 B=2 C=5 D=10

A with B (A returns)

A with C (A returns)

A with D

?

 

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no that would be 19mins, i’ll try again😚

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Posted (edited)

C with D (10 minutes)

D returns without his instrument (1 min)

B and A (2 minutes)

A returning (1 minutes)

A and D without his instrument (1 mins)

And you have 2 minutes to spare!

Edited by Sword
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Another old one:

A man is travelling with his dog, a rabbit and a large cabbage. He comes to a river and can only cross by the small boat tied up there. Unfortunately he can only fit one of his items in the boat with him, and he cannot leave the dog with the rabbit, or the rabbit with cabbage unattended at any time. How does he get across?

 

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Man takes rabbit across. 

Then goes back and take dog across. Take Rabbit on second journey back.

Leave Rabbit and take cabbage across. 

Go back and take Rabbit across. 

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7 hours ago, Sword said:

C with D (10 minutes)

D returns without his instrument (1 min)

B and A (2 minutes)

A returning (1 minutes)

A and D without his instrument (1 mins)

And you have 2 minutes to spare!

Not allowed! D takes 10 minutes there and 10 minutes back..

Try again.

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13 hours ago, Peckris 2 said:

It's actually even easier than that! You're right that the second man catches up with the first at 10:15. That's exactly two hours after he set out. The dog is running at 5mph, so in 2 hours he runs 10 miles.

That's how I did it.....

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10 hours ago, Peckris 2 said:

Not allowed! D takes 10 minutes there and 10 minutes back..

Try again.

The puzzle didn't say that each musician cannot part with their beloved instrument. 

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Posted (edited)
4 hours ago, Sword said:

The puzzle didn't say that each musician cannot part with their beloved instrument. 

As with most logic puzzles, the actual thing is the calculation - the rest is just window dressing to make it seem more fun, or to misdirect.

To restate :

Person 1 takes 1 minute to cross the bridge in either direction; Person 2 : 2 minutes, Person 3 : 5 minutes, Person 4 : 10 minutes. They can only cross in pairs (or singly) and have only 17 minutes to do it. (You also have to factor in that one of those who cross has to bring the torch back each time).

Edited by Peckris 2

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3 hours ago, Peckris 2 said:

As with most logic puzzles, the actual thing is the calculation - the rest is just window dressing to make it seem more fun, or to misdirect.

To restate :

Person 1 takes 1 minute to cross the bridge in either direction; Person 2 : 2 minutes, Person 3 : 5 minutes, Person 4 : 10 minutes. They can only cross in pairs (or singly) and have only 17 minutes to do it. (You also have to factor in that one of those who cross has to bring the torch back each time).

You snuck in the 'or singly' there so not exactly a 'restate'😂

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Got it (?) A=1 B=2 C=5 D=10

A with B (A returns) = 3mins

A remains and sends C with D (B returns)= 12mins

A with B = 2mins

The 'singly' was a red herring then, you little devil😈

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OK, here's a complex one.

Three logic sudents are sat in class in front of their professor. He shows them 8 stamps, 4 green and 4 red, and sticks 2 stamps on the foreheads of each student and places the other 2 face down on his desk. He then asks the first student if he knows which stamps are on his own forehead. The student thinks for a minute and says 'no.' The professor then asks the second student and again the student says 'no' to the same question. The professor then asks the third student the same question and gets the same answer. The professor then goes back to the first student and again poses the same question. The student thinks for a bit longer, but again says 'no' he doesn't know which stamps are on his forehead. Finally, the professor asks the second student the question again. The student thinks for some time and says 'yes' the stamps on my forehead are......?????

Each student can only see the stamps on the other students foreheads, students cannot remove stamps and there are no mirrors or other reflective surfaces in the room. In other words this has to be solved by logic alone. What stamps does the second student have on his forehead?   

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On 3/31/2020 at 8:27 PM, Sword said:

The poles need to be touching.

Those poles they get all the jobs in amazon

Thinking about it they can keep them!

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1 hour ago, DaveG38 said:

OK, here's a complex one.

Three logic sudents are sat in class in front of their professor. He shows them 8 stamps, 4 green and 4 red, and sticks 2 stamps on the foreheads of each student and places the other 2 face down on his desk. He then asks the first student if he knows which stamps are on his own forehead. The student thinks for a minute and says 'no.' The professor then asks the second student and again the student says 'no' to the same question. The professor then asks the third student the same question and gets the same answer. The professor then goes back to the first student and again poses the same question. The student thinks for a bit longer, but again says 'no' he doesn't know which stamps are on his forehead. Finally, the professor asks the second student the question again. The student thinks for some time and says 'yes' the stamps on my forehead are......?????

Each student can only see the stamps on the other students foreheads, students cannot remove stamps and there are no mirrors or other reflective surfaces in the room. In other words this has to be solved by logic alone. What stamps does the second student have on his forehead?   

... One of each.

On the first run through if any student can see four stamps the same colour they would know immediately that their own must both be the opposite colour. So each must be seeing at least one student with opposite colour stamps.

On the second run, the first student cannot be sure if his stamps are the same colour yet, so cannot answer. By the second student the only possibility remaining is that his stamps are opposite colours, otherwise the first student would have had the answer.

(I think!)

 

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2 hours ago, Diaconis said:

Got it (?) A=1 B=2 C=5 D=10

A with B (A returns) = 3mins

A remains and sends C with D (B returns)= 12mins

A with B = 2mins

The 'singly' was a red herring then, you little devil😈

Not quite - the 'singly' simply restates the fact that one of the musicians must bring the torch back for the next pair, i.e. returns across 'singly'.

But yes, you got the answer. 😊

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