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Answer fourd. Keep Riddle thread going

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Yeah. My brain only has like, 800 bytes of random access memory, and I'm at work right now so 204 of those bytes are occupied. I can't be expected to retain all these steps and still find it fun to solve.

Edit: It can be solved without binary. But binary makes it more ubergeeked.
 
Originally posted by: mobobuff
Yeah. My brain only has like, 800 bytes of random access memory, and I'm at work right now so 204 of those bytes are occupied. I can't be expected to retain all these steps and still find it fun to solve.

It's not very hard. Just convert the wine bottles into binary, and for 1000 bottles, you'll need 10 binary digits. Have each servant represent a digit so...

Servant 1 drinks all the wine who's binary number has xxxxxxxxx1
Servant 2 drinks all the wine who's binary number has xxxxxxxx1x
Servant 3 drinks all the wine who's binary number has xxxxxxx1xx
etc....

After 24 hours, you just look at the combination of who died and you can identify the bottle.
 
Originally posted by: mobobuff

Edit: It can be solved without binary. But binary makes it more ubergeeked.

It can be solved without binary, but binary is the easiest thing to explain it in.
 
Originally posted by: TuxDave
Originally posted by: mobobuff
Yeah. My brain only has like, 800 bytes of random access memory, and I'm at work right now so 204 of those bytes are occupied. I can't be expected to retain all these steps and still find it fun to solve.

It's not very hard. Just convert the wine bottles into binary, and for 1000 bottles, you'll need 10 binary digits. Have each servant represent a digit so...

Servant 1 drinks all the wine who's binary number has xxxxxxxxx1
Servant 2 drinks all the wine who's binary number has xxxxxxxx1x
Servant 3 drinks all the wine who's binary number has xxxxxxx1xx
etc....

After 24 hours, you just look at the combination of who died and you can identify the bottle.

zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
 
Fine... apparantly only computer people will see the simplicity of the answer. Next riddle! Requires alot less math.

There's 5050 coins in a treasure chest and 100 pirates. They put the numbers 1-100 in a hat and each of them draw a number out without replacement. That number represents how much money they'll get. At the end of the drawing, all the pirates who are unhappy with their number can toss it back in the hat and those pirates draw again. This continues until no one wants to toss their number in or only one person tosses it in.

If you were playing this game, when would you toss your number back in the hat and when will you not?

(NO MATH INVOLVED!)
 
No takers?

Hint: Consider at first the person who drew 100 and the person who drew 99. What's the 100 guy going to do? As a result... what will the 99 guy going to do?
 
Originally posted by: TuxDave
No takers?

Hint: Consider at first the person who drew 100 and the person who drew 99. What's the 100 guy going to do? As a result... what will the 99 guy going to do?

Assumes all of the pirates use the same logic. It doesn't allow for greed. And, it doesn't allow for hope in the face of frustration. Using what you posted, "what will the 99 guy do?", etc. only the 1 guy would return his number. But, what about the guy with 20 who would rather take the chance that the guy with 40 was stupid or greedy or both - because 20 is pretty much chickenfeed to him.
 
Originally posted by: DrPizza
Originally posted by: TuxDave
No takers?

Hint: Consider at first the person who drew 100 and the person who drew 99. What's the 100 guy going to do? As a result... what will the 99 guy going to do?

Assumes all of the pirates use the same logic. It doesn't allow for greed. And, it doesn't allow for hope in the face of frustration. Using what you posted, "what will the 99 guy do?", etc. only the 1 guy would return his number. But, what about the guy with 20 who would rather take the chance that the guy with 40 was stupid or greedy or both - because 20 is pretty much chickenfeed to him.

ding ding ding! teh winnar! But yeah, if you know that a couple pirates are greedy and stupid, then maybe your strategy will change. *thinks of another puzzle*
 
You have 10 glasses filled with pills, 9 contain placebo and one contain deadly poisonous pills. You also have a weigth. The placebo pills weigh 1 g and the deadly weigh 1.1 g. You're allowed to use the weight 1 time. How will you decide which glass contains the deadly pills?
 
Originally posted by: biostud666
You have 10 glasses filled with pills, 9 contain placebo and one contain deadly poisonous pills. You also have a weigth. The placebo pills weigh 1 g and the deadly weigh 1.1 g. You're allowed to use the weight 1 time. How will you decide which glass contains the deadly pills?

do you have to eat all the placebos or just 1?
 
Originally posted by: biostud666
You have 10 glasses filled with pills, 9 contain placebo and one contain deadly poisonous pills. You also have a weigth. The placebo pills weigh 1 g and the deadly weigh 1.1 g. You're allowed to use the weight 1 time. How will you decide which glass contains the deadly pills?

Grab 1 pill from glass 1, 2 pills from glass 2, 3 pills from glass 3.. etc....

Weigh it. Find how much over 55 grams it weighs and divide it by 0.1 to find which glass it came from.

*** DID NOT GOOGLE
 
Originally posted by: TuxDave
Originally posted by: biostud666
You have 10 glasses filled with pills, 9 contain placebo and one contain deadly poisonous pills. You also have a weigth. The placebo pills weigh 1 g and the deadly weigh 1.1 g. You're allowed to use the weight 1 time. How will you decide which glass contains the deadly pills?

Grab 1 pill from glass 1, 2 pills from glass 2, 3 pills from glass 3.. etc....

Weigh it. Find how much over 55 grams it weighs and divide it by 0.1 to find which glass it came from.

*** DID NOT GOOGLE

of course you didn't. Cause anyone else would have said if the weight ends in .1, it's vial 1. If .2 it's vial 2. If .3, it's vial 3... and if it's an even amount, it's vial 10. 🙂 😛
They wouldn't have reduced it to a mathematical formula
 
Originally posted by: DrPizza
Originally posted by: TuxDave
Originally posted by: biostud666
You have 10 glasses filled with pills, 9 contain placebo and one contain deadly poisonous pills. You also have a weigth. The placebo pills weigh 1 g and the deadly weigh 1.1 g. You're allowed to use the weight 1 time. How will you decide which glass contains the deadly pills?

Grab 1 pill from glass 1, 2 pills from glass 2, 3 pills from glass 3.. etc....

Weigh it. Find how much over 55 grams it weighs and divide it by 0.1 to find which glass it came from.

*** DID NOT GOOGLE

of course you didn't. Cause anyone else would have said if the weight ends in .1, it's vial 1. If .2 it's vial 2. If .3, it's vial 3... and if it's an even amount, it's vial 10. 🙂 😛
They wouldn't have reduced it to a mathematical formula

😱
 
* Stolen from my weekly newsletter....

You get a job and the boss offers you two possible salaries. All payments are in lump sums at the end of the year. You plan to stay for 6 years, what do you do?

A) $40,000 for your first year and an $8,000 raise every year afterwards
B) $20,000 for your first 6 months of works and a raise of $2,000 for every 6 months afterwards.

What should you choose?
 
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