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Dry
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Brussels
Sep 2000
time: 06:23
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A man wants to change the bronze number on his house door.
He talks about it with his neighbours and they say they also want to change. It is decided that he will do the purchase for all of them.
He spots at the local store nice bronze numbers but they cost $1 for a 1, $2 for a 2, $3 for a 3 ... and $10 for a 0.
He noticed that one of his neighbour, although having a lower house number, will pay a higher price than him, while the other neighbour, having a higher house number will pay a lower price than him.
Knowing that
- the street has 150 houses (75 on both sides),
- odd numbers on one side (1-149), even on the other (2-150)
- there is no missing house/number
- the three neighbours live on the same side of the street
What are the house numbers of each.
EDIT (after Ramo's answer): The first neighbour pays $1 less than him, the other $7 more... and there is no heading 0: no 001 but 1, no 017 but 17.
Last edited by Dry on 06-11-2002 at 16:00
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One_Brow
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quote:
Originally posted by CyberGnu
The trick is to reduce the problem to one wife, then two etc etc.
If there was only one unfaithful woman in town, everyone would know who it was except her husband. That poor guy would know that all other women are faithful, therefore when the mayor says there is one unfaithful woman, it has to be his wife. He would thus shoot her.
If there were two unfaithful women in town, everyone would know except their husbands, who thinks there are only one woman. After the mayors speech, he expects that the other husband would realize his wife is unfaithful, and kill her at midnight.
When he wakes up in the morning and sees that no woman has been killed, he must draw the conclusion that the other guy knows of another unfaithful woman, which must be his own wife.
The other guy reasons the exact same way, so the second night both wifes are shot.
If there are three wifes, it would take three nights before the three husbands realize that the two women they know of aren;t dead yet.
After a lot of counting, and possibly tying all wifes to lamposts to keep them from running from a certain death, the 40 husbands realize that there must be one more than the 39 wifes they know of, and erupt in an orgy of violence.
Good times. |
I'm sorry, but the induction does not apply here, because each man knows the exact (with the exception of his own wife) count of the number of unfaithful wives. We don't have an escalting potential count. Either the men can figure out their wife is unfaithful immediately, or they can't.
Every man with a faithful wife knows the possible number of unfaithful wives is 40 or 41. He knows that every man with a faithful wife knows of either 40 or 41. He knows that every man with an unfaithful wife knows of either 39 or 40.
Every man with a unfaithful wife knows teh possible number of unfaithful wives is 39 or 40. He knows that every man with a faithful wife knows of either 39 or 40. He knows that every man with an unfaithful wife knows of either 38 or 39.
You can't bootstrap the 38 inductivly to 37, because 37 is not in the possible range of initial values. An induction has to start with possible values. Basically, the men know too much to allow the induction to begin.
The correct answer is that, without further information, nothing happens.
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Dauphin
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Caught in a tuna net
Jan 1970
time: 05:23
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quote:
Originally posted by Dry
A man wants to change the bronze number on his house door.
He talks about it with his neighbours and they say they also want to change. It is decided that he will do the purchase for all of them.
He spots at the local store nice bronze numbers but they cost $1 for a 1, $2 for a 2, $3 for a 3 ... and $10 for a 0.
He noticed that one of his neighbour, although having a lower house number, will pay a higher price than him, while the other neighbour, having a higher house number will pay a lower price than him.
Knowing that
- the street has 150 houses (75 on both sides),
- odd numbers on one side (1-149), even on the other (2-150)
- there is no missing house/number
- the three neighbours live on the same side of the street
What are the house numbers of each.
EDIT (after Ramo's answer): The first neighbour pays $1 less than him, the other $7 more... and there is no heading 0: no 001 but 1, no 017 but 17. |
I take it they can't be next door neighbours, because otherwise the purchaser of door numbers (d1) cannot be paying more than the neighbour with a higher door number (d2) and less than the neighbour with a lower door number (d3).
If d1 < d3; then the digits for d3 must be smaller. i.e the final digit has recursed back to zero or one.
If d2 < d1; then the same logic applies.
This is not possible unless the ten's digit is different for d1, d2 & d3 - not obtainable when the difference between them is limited to 4.
Assuming non-consecutive neighbours. The easier way to increase the cost by $1 dollar is to add a prefix 1. The easiest way to decrease the cost by $7 is to increase a number by 2 (this increases the ten's digit by 1 and decreasing the unit digit by -8 +1 = -7).
The house numbers on the odd side are 9, 19, 21; 19, 29, 31; 29, 39, 41; etc or any number of configurations.
Edit - Just realised that I've worked it out as them both paying less than him. Plus some other silly errors. 
Last edited by Dauphin on 07-11-2002 at 00:04
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Dauphin
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Caught in a tuna net
Jan 1970
time: 05:23
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Reworking a bit of logic (i.e adding the prefix one, and subtracting two from the tens column to get a net reduction of one in the sum of digits, for the final figure):
29, 31, 111.
39, 41, 121.
49, 51, 131.
59, 61, 141.
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One_Brow
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quote:
Originally posted by Sagacious Dolphin
I take it they can't be next door neighbours, because otherwise the purchaser of door numbers (d1) cannot be paying more than the neighbour with a higher door number (d2) and less than the neighbour with a lower door number (d3). |
Although this wasn't the answer to the problem, there is one value of x where p(x) > p(x+2) > p(x+4) (p(x) being the price of x).
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rah
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Apolyton Prince of Moderators, Master of Reason
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Lord of the Ferrets
Jan 1970
time: 23:23
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quote:
Originally posted by Dry
The answer is: impossible
So there must be a trick...  9 and 6 are same!!!
So, the answer is: 87, 89, 91 |
Naw, I actually took a couple of minutes and set up a spreadsheet to calculate it. Upon checking it, I thought it was impossible also, but then after reading Sagacious Dolphin's answer, I went back and looked at it an saw it. I must have missed it originally.
108 110 112 is correct.
Cost 19, 12, 4.
It was the only one that worked.
The funny thing was, that I figured it would have to happen on the even number side of the street and it would be somewhere between 98 and 120, and still didn't see it when it was staring at me. So I went ahead and calculated it for the odd numbers. 
RAH
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The Vagabond
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of realpolitik and counterpropaganda
Jan 1970
time: 05:23
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A group of four persons (A, B, C and D) needs to cross a bridge. It takes person A 1 min to cross the bridge, person B -- 2 min, person C -- 4 min, person D -- 5 min. One of the difficulties they face is that the bridge is too shabby and narrow, so that maximum two persons can be crossing it at a time. Another problem is that it happens at night, and they need a lantern. But there is only one lantern for the whole group. Therefore, while necessary, someone will have to keep returning back in order to bring the lantern to the remainder of the group.
Question: How should they organize the crossing to do it as fast as possible? How much time will it take them?
Note: When two persons cross the bridge, it is implied that they travel at the speed of the slowest person, of course.
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JohnM2433
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Milwaukee, Wisconsin
Jul 2002
time: 21:23
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*bump*
Isn't anyone going to post another one?
I loved this thread! I don't want it to die! Noooo.....
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Just wanted to say that JohnM is right about the dividing of the land beetween sons.
Five is impossible, four is trivial.
The four Colour theorem, proven rather recently (in mathematical terms), states that a planar map (thats more precise than just 2D since the surface of the Earth is 2D for example) can never need more than 4 colours to color as in loinburgers question (meaning no same colour touching, not including corners etc) and was the first proof ever in mathematical history that was not checkable by hand (they used computers to do a divide and conquer method, meaning breaking all the possible cases in different categories which could each be solved) and thus is sorta classical and often referred to.
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06-11-2002 15:12
).
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