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chadtm80

Riddle me this Thread

#1 Postby chadtm80 » Thu Dec 02, 2004 12:42 pm

First Riddle to Solve

What row of numbers comes next?

1
11
21
1211
111221
312211
13112221
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Re: Riddle me this Thread

#2 Postby stormraiser » Thu Dec 02, 2004 12:46 pm

chadtm80 wrote:First Riddle to Solve

What row of numbers comes next?

1
11
21
1211
111221
312211
13112221


1113213211?
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#3 Postby CaptinCrunch » Thu Dec 02, 2004 1:00 pm

Heres one for the masses.

3 men working out of town need a motel room for the night, they pull into a Days Inn and ask how much is a room for the night? the bellhop says $30 dollars. The 3 men agree to pay $10 each for the room, later the clerk realizes that the men were charged $5 dollars to much and gave the bellhop $5 to take back up to the men. The bellhop desides that he will keep $2 dollars and give each of the men back $1 dollar each.

So now the men have paid $9 dollars each for the room thats $27 dollars total and the bellhop keep $2 dollars which makes $29 dollars total (9+9+9=27+2=29). wheres the other dollar???
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#4 Postby chadtm80 » Thu Dec 02, 2004 1:03 pm

The next row is indeed
1113213211

Starting with the second line, every line describes the line before it. In writing, it is:
One One
Two Ones
One Two One One
etc.
etc.

Great Job Storm.. You got it a heck of alot faster then I did..


Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?
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#5 Postby therock1811 » Thu Dec 02, 2004 1:20 pm

One hasn't been born yet?
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#6 Postby CaptinCrunch » Thu Dec 02, 2004 1:22 pm

therock1811 wrote:One hasn't been born yet?



thats what I was going to say :lol:
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#7 Postby The Big Dog » Thu Dec 02, 2004 1:35 pm

chadtm80 wrote:Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?

"Them" refers to the woman and the 7 children. Total of 8. Half of them are boys = 4. Three daughters, plus the woman = the other 4.
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#8 Postby GalvestonDuck » Thu Dec 02, 2004 1:39 pm

CaptinCrunch wrote:Heres one for the masses.

3 men working out of town need a motel room for the night, they pull into a Days Inn and ask how much is a room for the night? the bellhop says $30 dollars. The 3 men agree to pay $10 each for the room, later the clerk realizes that the men were charged $5 dollars to much and gave the bellhop $5 to take back up to the men. The bellhop desides that he will keep $2 dollars and give each of the men back $1 dollar each.

So now the men have paid $9 dollars each for the room thats $27 dollars total and the bellhop keep $2 dollars which makes $29 dollars total (9+9+9=27+2=29). wheres the other dollar???



If the hotel room was actually $25 and the overpaid $5 got split this way -- $1 to each man (for a total of $3) and $2 was kept by the bellhop ($25+$3+$2=$30). There is no extra dollar. Furthermore, each man paid $9 not for the cost of the ROOM, but for the cost of the ROOM and the THEFT by the bellhop. The cost of the room is $25. The bellhop kept $2. $25+$2=$27 $27 divided by the three men is $9, which is what each man paid. $27 + the $3 refunded to them = $30.

:)
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#9 Postby CaptinCrunch » Thu Dec 02, 2004 1:44 pm

GalvestonDuck wrote:
CaptinCrunch wrote:Heres one for the masses.

3 men working out of town need a motel room for the night, they pull into a Days Inn and ask how much is a room for the night? the bellhop says $30 dollars. The 3 men agree to pay $10 each for the room, later the clerk realizes that the men were charged $5 dollars to much and gave the bellhop $5 to take back up to the men. The bellhop desides that he will keep $2 dollars and give each of the men back $1 dollar each.

So now the men have paid $9 dollars each for the room thats $27 dollars total and the bellhop keep $2 dollars which makes $29 dollars total (9+9+9=27+2=29). wheres the other dollar???



If the hotel room was actually $25 and the overpaid $5 got split this way -- $1 to each man (for a total of $3) and $2 was kept by the bellhop ($25+$3+$2=$30). There is no extra dollar. Furthermore, each man paid $9 not for the cost of the ROOM, but for the cost of the ROOM and the THEFT by the bellhop. The cost of the room is $25. The bellhop kept $2. $25+$2=$27 $27 divided by the three men is $9, which is what each man paid. $27 + the $3 refunded to them = $30.

:)


very good!!:)
It took 2 hours before someone here at work got it :)
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#10 Postby GalvestonDuck » Thu Dec 02, 2004 2:00 pm

Thank ya! :)

It throws people because they try to add in the $2 for the bellhop AFTER figuring what the men paid, when in fact, part of what the men paid INCLUDES the $2. :)
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#11 Postby chadtm80 » Thu Dec 02, 2004 2:13 pm

The Big Dog wrote:
chadtm80 wrote:Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?

"Them" refers to the woman and the 7 children. Total of 8. Half of them are boys = 4. Three daughters, plus the woman = the other 4.

Sorry.. Nope.. Logical, but not the answer
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#12 Postby JenBayles » Thu Dec 02, 2004 2:17 pm

chadtm80 wrote:
The Big Dog wrote:
chadtm80 wrote:Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?

"Them" refers to the woman and the 7 children. Total of 8. Half of them are boys = 4. Three daughters, plus the woman = the other 4.

Sorry.. Nope.. Logical, but not the answer

One is androgynous?!
:)
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#13 Postby GalvestonDuck » Thu Dec 02, 2004 2:20 pm

chadtm80 wrote: Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?


So's the other half. :) They're ALL boys.
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#14 Postby JenBayles » Thu Dec 02, 2004 3:17 pm

GalvestonDuck wrote:
chadtm80 wrote: Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?


So's the other half. :) They're ALL boys.


Oi! I knew I shouldn't have even opened up this thread with this hangover.
:Pick:
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#15 Postby The Big Dog » Thu Dec 02, 2004 4:25 pm

JenBayles wrote:
chadtm80 wrote:
The Big Dog wrote:
chadtm80 wrote:Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?

"Them" refers to the woman and the 7 children. Total of 8. Half of them are boys = 4. Three daughters, plus the woman = the other 4.

Sorry.. Nope.. Logical, but not the answer

One is androgynous?!
:)

I think you mean hermaphroditic.

Or not. :-)
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#16 Postby yoda » Thu Dec 02, 2004 4:26 pm

GalvestonDuck wrote:
chadtm80 wrote: Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?


So's the other half. :) They're ALL boys.


I would agree with that Duckie.
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#17 Postby JenBayles » Thu Dec 02, 2004 4:50 pm

The Big Dog wrote:
JenBayles wrote:
chadtm80 wrote:
The Big Dog wrote:
chadtm80 wrote:Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?

"Them" refers to the woman and the 7 children. Total of 8. Half of them are boys = 4. Three daughters, plus the woman = the other 4.

Sorry.. Nope.. Logical, but not the answer

One is androgynous?!
:)

I think you mean hermaphroditic.

Or not. :-)


Yeah! Yeah! That's it! <removes head from rear end>
:A:
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#18 Postby chadtm80 » Fri Dec 03, 2004 11:48 am

GalvestonDuck wrote:
chadtm80 wrote: Another One: A lil eaiser I believe

A woman has 7 children, half of them are boys.
How can this be possible?


So's the other half. :) They're ALL boys.


Ding Ding Ding.. You got it Duckie.. There all boys.. There for Half of them are boys :-)

New One:

You throw away the outside and cook the inside. Then you eat the outside and throw away the inside. What did you eat?
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#19 Postby JenBayles » Fri Dec 03, 2004 11:52 am

chadtm80 wrote: New One:

You throw away the outside and cook the inside. Then you eat the outside and throw away the inside. What did you eat?


Haggis? :lol:
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#20 Postby chadtm80 » Fri Dec 03, 2004 2:17 pm

No.. LOL
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