L A T E R A L
That is thinking.
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You can’t trust cretins with logic.
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Nor Cretians. They lie!
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This is a lie.
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In all seriousness I took an entire class on this line of reasoning in college. We never got to any real conclusions but one of the things that stuck with me is metalanguage and levels of meaning. For instance: Tuesday is a day of the week.
Tuesday has seven letters.
Despite being the same word, Tuesday has two different meanings. The semantic meaning points to a day of the week. The metalanguage meaning points to the word object itself, the seven letters of “Tuesday.”
Then there was some argument I didn’t follow and can’t remember explaining how these unresolvable paradoxes are really a clash of language and metalanguage, and the language is assigned one truth value while the metalanguage is assigned the opposite.
And we, silly humans, mash the two different objects together because they arise from the same letters as written.
3 Likes
The person was hit by a car outside of a place they produce currency. The license plate of the vehicle tore off in the impact.
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The only American mint I know is in Kentucky
You both have the correct kind of mint, but not the right kind of plate
Is there a bullet-hole in the deceased?
Yes
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In that case, the person was stealing a plate used to print money with the intent of creating counterfeit bills and was shot by security trying to escape.
This was my actual first instinct, should have gone with it, right or wrong.
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The only riddle that comes to mind right now is one I made up in Japanese, and it’s awful. I will just write the question and answer here, so anyone here who might have dabbled in Japanese might be able to reverse engineer it for a groan, at best.
“What sea creature is bigger than a whale?”
“A dolphin?”
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Well, I’ve got one more and then I’ll leave this be unless the community wants to keep it going.
Most of you have probably done the easier version of this question:
You have 8 bb’s (ball bearings, little steel spheres) and one of them is slightly heavier than the other 7. Your only tool is a balance scale. Can you find the heavy bb, only using the scale twice?
The real question:
OK, now you have 12 bb’s. One of them is rogue, either heavier OR lighter, but you don’t know which. It just weighs different. How many times do you need to use the balance scale to find the rogue bb?
This is amazing and I don’t want the translation.
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I’ve seen these types of questions around before but never bothered to figure them out. Let’s have a go.
On one side of the scale, place BBs 1, 2, and 3. On the other side, place 4, 5, and 6.
If the (1st measurement) scale is balanced, remove and disregard BBs 1-6. Place BB 7 one side and BB 8 on the other to determine which is the heavier.
If the (1st measurement) scale is not balanced, the BBs from the heavier side are now a, b, and c. Place a on one side of the scale and b on the other and do the 2nd measurement.
If the (2nd measurement) scale is balanced, BB c is the heaviest. Otherwise, the BB on the heavier side of the scale is the heaviest.
Let me work on the second part for a bit…
3 Likes
Four, I think. Details later, if necessary.
Answer
Concurred: 4
Procedure
Place BBs 1, 2, 3, and 4 on one side of the scale (group A), and BBs 5, 6, 7, and 8 on the other side (group B). Make note of which side is heavier or that it is balanced (1st measurement).
Retain BBs group A on one side and place BBs 9, 10, 11, and 12 on the other side, group C. Make note of which side is heavier or that is is balanced (2nd measurement).
If group A was balanced with group B, group C contains the rogue BB, and the ‘lighter’ or ‘heavier’ status should be known. If group A was balanced with group C, then group B contains the rogue BB and its ‘lighter’ or ‘heavier’ status should be known. If both group A-group B and group A-group C were unbalanced, then group A contains the rogue BB.
The group containing the rogue BB is all that is important from here out, and we will call it group R regardless of its prior identity.
Split group R into two subgroups of 2, group R1 and group R2. If the rogue BB is known to be heavier, we will be looking for the heavier of these two subgroups, heavy. If the rogue BB is known to be lighter, we will be looking for the lighter of the two subgroups, light. Place group R1 and group R2 on either side of the scale (3rd measurement). Note which is either heavier or lighter, depending on the criterion identified above. Take the heavier or lighter of the two subgroups and split it into individual BBs and place them on either side of the scale (4th measurement). The rogue BB is identified based on whichever side is heavy or light.
Bingo! 
3 Likes
“I have eaten over a hundred humans!” boasted the ogre.
“Surely it’s not that many!” countered a princess.
“It’s definitely more than one!” snapped her brother.
If only 1 of these characters is accurate, how many humans has the ogre eaten?
What is heavy when written forwards, but not when it’s written backwards?
What man dies if he is inside a house?
The more you take away, the bigger I become. What am I?
The princess and her brother?
skip
skip
a hole
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You got the “hole” correct. 
Less than 1
Reasoning
If the ogre is right, then so is the brother.
If the brother is right, then the princess is right for any value between 1 and 100, and then the ogre is right for any value that’s over 100.
Leaving any value less than 1 as only matching the princess’ statement.
1 Like