Riddle me this, Riddle me that!

I tried looking for the 'real' answer, but the website I borrowed this from doesn't post answers.

I tried the riddle myself, and got a different answer than StaggerLee, I believe there's a way to work around the 4-7 issue.
 
I tried looking for the 'real' answer, but the website I borrowed this from doesn't post answers.

I tried the riddle myself, and got a different answer than StaggerLee, I believe there's a way to work around the 4-7 issue.

Google "German Block calender puzzle"?

My final answer is
Block A: 0, 1, 2, 3, 4, 5
Block B: 0, 1, 2, 6, 7, 8
6 rotates 180 to form a 9
 
Google "German Block calender puzzle"?

My final answer is
Block A: 0, 1, 2, 3, 4, 5
Block B: 0, 1, 2, 6, 7, 8
6 rotates 180 to form a 9

Yes, that's the answer I got!
The double 3 in StaggerLee's answer isn't needed.

No, google is NOT your friend! Leave him alone!

I suppose you can post that riddle of yours again. I was about to post an answer when I noticed my quoted text changed:wtf:
 
Jack and Mike are out chilling when Jack says, "I bet you R500 I can jump from America to China."
Mike keen for Easy money agrees and soon loses, how did Jack win?
 
Jack and Mike are out chilling when Jack says, "I bet you R500 I can jump from America to China."
Mike keen for Easy money agrees and soon loses, how did Jack win?

They were standing outside the Chinese Embassy in the USA and then Jack jumps from the USA into China.

Why they were using Rand as currency is a bit of a mystery.
 
Exactly, your up

Oh, that's unexpected:)

Hope this one hasn't bee posted yet.

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?
 
Ok if the first guy switches the left switch down then he can keep track of people going through. Then after that when a prisoner enters for the first time he flicks the switch on the left to ON then on any following trips he plays with the other switch. If the switch is already at ON the he plays the right switch until its open for him to press.

This means only prisoner 1 can reset the left switch to OFF and each of the other 22 prisoners only turn it to ON once ever. When prisoner 1 resets 22 switches then he knows everyone has been through once for sure. The right switch is purely for fun.
 
Ok if the first guy switches the left switch down then he can keep track of people going through. Then after that when a prisoner enters for the first time he flicks the switch on the left to ON then on any following trips he plays with the other switch. If the switch is already at ON the he plays the right switch until its open for him to press.

This means only prisoner 1 can reset the left switch to OFF and each of the other 22 prisoners only turn it to ON once ever. When prisoner 1 resets 22 switches then he knows everyone has been through once for sure. The right switch is purely for fun.

At first I didn't agree with your answer, but after reading it 3 or 4 times, I see that your solution will work:)
Well done, you're up!
 
Wow sweet, traffic has its uses after all.

A young scholar grew tired of hearing how knowledgeable the Emperors Advisor Richard is. One day he decided to challenge the advisor and try knock him down

He gave the advisor 2 options, 100 easy questions or 1 difficult questions. The advisor, being a busy man, decided on the hard question.

The Scholar then asked "Which came first, the chicken or the egg?"
"Simple, the Egg" replied Richard. The scholar seeing an opening asked him why.

What did the Advisor, Richard say to end the discussion and left the scholar amazed?
 
Wow sweet, traffic has its uses after all.

A young scholar grew tired of hearing how knowledgeable the Emperors Advisor Richard is. One day he decided to challenge the advisor and try knock him down

He gave the advisor 2 options, 100 easy questions or 1 difficult questions. The advisor, being a busy man, decided on the hard question.

The Scholar then asked "Which came first, the chicken or the egg?"
"Simple, the Egg" replied Richard. The scholar seeing an opening asked him why.

What did the Advisor, Richard say to end the discussion and left the scholar amazed?

Wasn't he only supposed to ask one question? "why?" is a second question. That wouldn't amaze the scholar though, so I'm not sure.
 
Wasn't he only supposed to ask one question? "why?" is a second question. That wouldn't amaze the scholar though, so I'm not sure.

Amazed is a strong description (the original used the same ending) but your right, there only is one question agreed to so "Why?" won't count.

Your up Riddle master
 
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Hint : Count the letters and try splitting the letters up into groups.
 
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