Tricky Puzzles - Topic 3

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VirtLands
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brain food

Post by VirtLands » Tue Jun 11, 2013 8:03 pm

yot yot5 wrote:I've spent a long time on this... I hope you're ready for some REAL brain food!
CAPAIN SCRITTERBRAIN'S TREASURE
This puzzle was originally found in an Usborne Superpuzzles book,
I certainly "eyed" it, and tried thinking about it.
The puzzle seems to be based on a map system,
but I couldn't visualize the map because I don't have the map in front of me.
Has anyone else gotten it? Maybe some coordinates would help.
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Re: brain food

Post by yot yot5 » Tue Jun 11, 2013 8:22 pm

VirtLands wrote:
yot yot5 wrote:I've spent a long time on this... I hope you're ready for some REAL brain food!
CAPAIN SCRITTERBRAIN'S TREASURE
This puzzle was originally found in an Usborne Superpuzzles book,
I certainly "eyed" it, and tried thinking about it.
The puzzle seems to be based on a map system,
but I couldn't visualize the map because I don't have the map in front of me.
Has anyone else gotten it? Maybe some coordinates would help.
Captain Scritterbrain from the original post wrote:You might've noticed that the locations were placed in order, from WEST to EAST. The first five locations were on the west side of Wonderland, and the last five locations were on the east side of Wonderland.
If you don't play Wonderland often, you may not know that the Fire Island Acid Pools, Fire Island Jungle, and Sundog Island aren't on Wonderland's mainland. All the other info should be in the puzzle.
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VirtLands
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math puzzles

Post by VirtLands » Tue Jun 11, 2013 8:52 pm

Okay, now I shall give an invented math quiz.
Some of these may be easy, others not....

(Please give the answers in radians, and assume all is in radians,...)
;; arcsine =inverse sine,
;; i = √-1

All of the following have answers, even if your calculator can't do it. :twisted:

(A) find arcsine(0.5)
(B) find arcsine(2)
(C) find cosine(2+3i)
(D) find arcsine(1+i)

(E) In which base (of number systemₓ) does 10ₓ x 11ₓ = 12ₓ ?
-----------------------------------------------------------------------------------

(F) A Triangle exists as follows, with leg measurements as shown: (figure is NOT proportional)
Image
b=2
c=3
a=10


(..In other words, leg a is longer than b+c combined, and all lines are straight lines. :shock: )

Please find the angles α, β and show it in degrees.
Optionally, if you can, calculate the AREA of this impossible triangle.

(Clue: These are imaginary angles that cannot exist in our real world;
The triangle itself is so impossible that it can't even be drawn,
yet there is an imaginary answer. )

The best way to even start to solve this would be to use the Law of Cosines:
http://en.wikipedia.org/wiki/Law_of_cosines

Image
-------------------------------------------------------------------------------
(update.)
Day 2 has come and gone, and no one has solved any of these. (A)--(F)

I'm not surprised, those math puzzles are hard for me to solve as well.
After a week, I'll be back, and post the "answers" . :)
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Post by Muzozavr » Thu Jun 13, 2013 6:25 pm

A: pi/6 or 5pi/6.

More exactly, it's pi/6+2npi (n = whole number) because sine is a periodic function.

This is the only one I can do, because it's a real angle. :roll: :oops: :lol:

The imaginary ones? I have just finished the last grade high school (more specifically, our equivalent of it) and we have never studied imaginary numbers and I have no idea how to even begin working with them. I can imagine them, certainly (that's why they're called imaginary :wink:) but I still don't know how to actually work with them in any scenario that could be considered even remotely complicated. :?

We haven't studied non-decimal numbering systems, either, but E looks like it could be plausibly solved with a few hours of Wikipedia surfing.

Your imaginary triangle looks like an epic concept for a not-very-obvious spatial anomaly. You could just walk around that and never even notice that such a triangle can't exist... :shock:

If I ever end up writing some supernatural/mystical/paranormal/whatever story, mind if I use it? I'll give proper credits, of course.
Rest in peace, Kym. I hardly knew ya.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
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math quiz answers _A

Post by VirtLands » Fri Jun 14, 2013 2:25 am

Muzozavr wrote:Your imaginary triangle looks like an epic concept for a not-very-obvious spatial anomaly.
You could just walk around that and never even notice that such a triangle can't exist... :shock:
If I ever end up writing some supernatural/mystical/paranormal/whatever story, mind if I use it?

Sure, it would be a pleasure, to have someone write about unusual math figures.
-----------------------------------------------------------------------------------

I'll go ahead and show the answers for (A)--(F)

(A) find arcsine(0.5) :
This one is the easiest to solve, you can use your calculator for this one:

arcsine(0.5) = 30° = pi/6 ≈ 0.523598776 + 2kpi (where k= any integer)

http://www.wolframalpha.com/input/?i=ar ... 9&dataset=

-----------------------------------------------------------------------------------
(B) find arcsine(2) : (This is going to be hard, take an hour to solve :shock:)

The answer can only be a complex (number) angle 'x',
since no real angle by itself can have its sine =2

http://en.wikipedia.org/wiki/Complex_number

Our goal is to find angle x, where x =(a+ib) = arcsine(2)

i=√-1 ;; the red-hot i is the square root of -1

2 = sin(x) = sin(a+ib)

;; some useful Sum of Angle Identities:
;; sin (α+β) = (sin α)•(cos β) + (cos α)•(sin β)
;; cos (α+β) = (cos α)•(cos β) – (sin α)•(sin β)

2 = sin(a+ib) = sin(a)•cos(ib) + cos(a)•sin(ib)

;; some relations between Hyperbolic and Trigonometric Identities
;; reference --
;; sinh(α) = -i•sin(iα)
;; cosh(α) = cos(iα)
;;
;; -sinh(α) = i•sin(iα)
;; -i•sinh(α) = -sin(iα)
;; i•sinh(α) = sin(iα)
;; -sinh(iα) = i•sin(-α)
;; -sinh(-iα) = i•sin(α)
;; -i•sinh(-iα) = -sin(α)
;; i•sinh(-iα) = sin(α)


2 = sine( a+ib ) = sin(a)•cos(ib) + sin(ib)•cos(a)
2 = sine( a+ib ) = sin(a)•cosh(b) + sinh(b)•cos(a) ;; substitutions

Let's try cos(a)=0, a=90° = ½pi, and therefore cos(a)=0.0, sin(a)=1.0

2 = sine( a+ib ) = sin(a)•cosh(b) ;; simplifications
2 = sine( a+ib ) = cosh(b)
2 = cosh(b)
b = arccosh(2), ;; find the inverse hyperbolic cosine of 2
b ≈1.3169578.. = ln(2+√3) ;; (natural log)

since a =90° = pi/2 rad, then

2 = sin( a + ib ) = sin( ½pi + i•1.3169578 ) = sin(x)

We found our answer, angle x = ½pi + i•1.3169578

x = 1.570796327 + i•1.3169578 ;; = a complex number

or, in higher precision:

x = 1.570796326794896619231321691639751442098584699687552
+ i•1.316957896924816708625046347307968444026981971467516
(shown in radians)

x = (90 -75.46i )° ;;

sin(90° -75.46i°) = 2 ;; (<-- the answer in degrees °)

Wolfram Alpha result: http://www.wolframalpha.com/input/?i=arcsine%282%29
------------------------------------------------------------------------------------

(C) find cosine(2+3i)

Use the sum of angles formula:

;; cos (&#945;+&#946;) = (cos &#945;)•(cos &#946;) – (sin &#945;)•(sin &#946;)

cos(2 + 3i) = (cos 2)(cos 3i) -(sin 2)(sin 3i)
cos(2 + 3i) = (cos 2)(cosh 3) -i•(sin 2)(sinh 3) ;; conveniently replace trig values with hyperbolic values
cos(2 + 3i) &#8776; -4.1896256 -i•9.1092278

as proven on Wolfram Alpha: http://www.wolframalpha.com/input/?i=cosine%282%2B3i%29+

------------------------------------------------------------------------------------
(D) find arcsine(1+i) :::::: I'm exhausted, shall skip (D) and maybe solve it later. :oops:
------------------------------------------------------------------------------------

(E) In which base (of number system&#8339;) does 10&#8339; x 11&#8339; = 12&#8339; ? :)

Most number systems are graduations of powers, in this kind of fashion:::

a*x^4 + b*x^3 + c*x^2 + d*x^1 + e*x^0

where a..e may 0,1,.. or any other integer place holder value.

let's find the Base... (Base)

10&#8339; x 11&#8339; = 12&#8339;

(1•B +0) x (1•B +1) = (1•B +2)

(B)(B+1) = (B+2)

B² + B = B+2
B² = 2
B = ±&#8730;2 &#8776; ±1.414213562

There's our answer, a number system with an irrational base of ±&#8730;2

would allow 10 x 11 = 12,

and the proof:

(±&#8730;2•1 + 0) x (±&#8730;2•1 + 1) = (±&#8730;2•1 + 2)
±&#8730;2(&#8730;2 +1) = ±&#8730;2 +2

I know, that was too weird, sorry.
--------------------------------------------------------------------------------------

(F) A Triangle exists as follows, with leg measurements as shown: (figure is NOT proportional)
Image
b=2, c=3, a=10

So, let's use the Law of Cosines to solve this impossible triangle: :lol:

http://www.mathopenref.com/lawofcosines.html
http://en.wikipedia.org/wiki/Law_of_cosines

first, lets find angle (&#945;) alpha, computed from legs a,b,c:

&#945; = arccos[ (b² + c² - a²)/(2bc) ]

&#945; = arccos[ (2² + 3² - 10²)/(2•2•3) ]

&#945; = arccos[ -87/12 ] = arccos[ -29/4 ] = arccos[ -7.25 ]

;; can do some substitution if we wished to, but it's not necessary
http://www.wolframalpha.com/input/?i=arccos%28-a%29

Let's find arccos( -7.25 ), ....

Can do a Wolfram Alpha lookup, to see what arccos( -7.25 ) is:

http://www.wolframalpha.com/input/?i=ar ... +-7.25+%29

{ I can go through all the painful details on how to slowly compute arccos( -7.25 )
as in math puzzle (B), or, I can get it straight from Wolfram. :exhausted: Image }

arccos( -7.25 ) = 3.14159 - i*2.66936 (in radians)

angle &#945; = (180 -i•152.9)° ;; <-- Our Answer shown in degrees. :)

___ So, there you have it, angle &#945; is a complex (imaginary angle); That is really mind-bending...

Next, use similar techniques to find angles &#946;, &#947;, (&#946;=Beta, &#947;=Gamma) ::

&#946; = arccos[ (a² + c² - b²)/(2ac) ]
&#947; = arccos[ (a² + b² - c²)/(2ab) ]

Solve those equations by substituting in the numbers ::

&#946; = arccos[ (10² + 3² - 2²)/(2•3•10) ] = arccos( 7/4 )
&#947; = arccos[ (10² + 2² - 3²)/(2•2•10 ] = arccos( 19/8 )

;; Look on Wolfram Alpha to get quick results....
;; ..results for angles &#946; & &#947; will both be imaginary :cool:

&#946; = arccos( 1.750 ) = i•1.1588103604
&#947; = arccos( 2.375 ) = i•1.5105477504
&#945; = arccos( -7.25 ) = 3.14159 - i•2.66936

In summary, all the angles of this triangle are imaginary, with angle &#945; being both imaginary and complex.

But, with this complex imaginary triangle, do the 3 angles still add up to
180°, which is traditional with Euclidean geometry ??

http://www.cut-the-knot.org/triangle/py ... ngle.shtml

Is &#945;+&#946;+&#947; = 180 = pi radians ? :shock:

I just checked it, and it's true, though none of the angles are real, they still add up to 180°

--------------------

Next, let's find the Area of the Triangle, using Heron's formula:

http://www.mathopenref.com/heronsformula.html

Image where p = semi-perimeter = (a+b+c)/2

p = 15/2 = 7.5

Area = &#8730;-464.0625 = i•(15/4)•&#8730;33

Area = &#8776; i•21.54210 cm² ( We can assume it's square centimeters. :) )

So, in this case, not only are all the angles imaginary, the area is also imaginary, given that it is a multiple of i=&#8730;-1, and
the legs are impossible too: b=2, c=3, a=10

That is all. Hope it was entertaining. That is as close to paranormal as I've been.
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Post by Muzozavr » Fri Jun 14, 2013 5:30 am

That was a fascinating read. Some of them I would've probably figured out the trick for, though I wouldn't have been able to carry the calculations all the way through. I was too busy trial-and-erroring specific numbering systems on Wikipedia, though, so I'd never solve E.
There's our answer, a number system with an irrational base of ±&#8730;2
You monster. :lol: Seriously, that was a beautiful piece of trickery, but I don't think there's anyone on our forum "mathy" enough to be able to solve such puzzles. :lol: I like this one.
Rest in peace, Kym. I hardly knew ya.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
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Post by yot yot5 » Fri Jun 14, 2013 8:21 am

Whoa, that was incredible. I didn't understand half of it, but the other half was really interesting to read. :)
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Post by yot yot5 » Thu Jun 20, 2013 7:59 pm

Since nobody seems to have solved the Captain Scritterbrain's Treasure puzzle, I decided to piece together the full solution.
Captain Scritterbrain wrote:1: Let's begin the fireflower trap. From clue 10, we know that the fireflower trap is not on the same side of Wonderland as the boulder trap, but only if the fireflower trap is on the west side of Wonderland. Imagine the fireflower trap was on the west side of Wonderland. The boulder trap WOULD NOT be on the same side, so it would be on the east side. Now imagine if the fireflower trap was on the east side of Wonderland. The boulder trap WOULD be on the same side, so it would be on the east side. This means that the boulder trap is on the east side of Wonderland, no matter where the fireflower trap is.

2: Now look at the exploding barrel trap. From clue 2, we know that it's not on the same side of Wonderland as the boulder trap, so it must be on the west side. From clue 1, we also know that it's not on Wonderland's mainland. The only location on the west side of Wonderland that's not on the mainland is Sundog Island. This means that the exploding barrel trap must be on Sundog Island.

3: Let's narrow down the possible locations for the fireflower trap. From clue 4, we know that the fireflower trap in a location inhabited by two species, and we know that Sundog Island already has the exploding barrels. This means that the fireflower trap must be in either Crystal Lake, Foggy Peak, or the Lonely Top.

4: Let's narrow down the possible locations for the cage trap. From clue 11, we know that the cage trap is somewhere in the foggy mountains, but NOT if the fireflower trap is on the west side of Wonderland. We know that that fireflower trap is in either Crystal Lake, Foggy Peak, or the Lonely Top, and all three of those locations are on the east side of Wonderland. This means that the cage trap is somewhere in the foggy mountains. (either Crystal Lake or Foggy Peak)

5: Let's narrow down the possible locations for the chomper trap. From clue 9, we know that the fireflower trap is not on the same side of Wonderland as the chomper trap, but only if the chomper trap is not on Wonderland's mainland. Imagine the chomper trap was not on Wonderland's mainland. It WOULD NOT be on the same side as the fireflower trap, so it must be on the west side. Hang on! On the west side of Wonderland, the only location not on Wonderland's mainland is Sundog Island, and that's already taken by the exploding barrels! That means that the chomper trap IS on Wonderland's mainland, and IS on the same side of Wonderland as the fireflower trap. This means it's in either Crystal Lake, Foggy Peak, or the Lonely Top.

6: From clue 8, we know that the fireflower trap is in a location which chompers have inhabited, but only if the swinging axe trap is in a very cold location. We know that Sundog Island is already taken by the exploding barrel trap. We also know that Crystal Lake, Foggy Peak, and the Lonely Top must (in some order) be hiding the fireflower trap, the cage trap, or the chomper trap. This means that the swinging axe trap IS NOT in a very cold location, which means the fireflower trap is in a location inhabited by chompers. The only location out of Crystal Lake, Foggy Peak, and the Lonely Top which is inhabited by chompers is Foggy Peak. This means that the fireflower trap must be in Foggy Peak.

7: Now that we know the fireflower trap is in Foggy Peak, the only possible remaining location for the cage trap is Crystal Lake. Now that we know the cage trap is at Crystal Lake, the only possible remaining location for the chomper trap is Lonely Top. We now know the locations of 4/10 traps.

8: From clue 3, we know that the firepit trap is in a location which scritters have inhabited. Out of the six remaining locations, only the Windy Hills and the Fire Island Jungle are inhabited by scritters. Both of these locations are 30 square stinkomiles large, so the firepit trap is in a location 30 square stinkomiles large.

9: From clue 5, we know that the swinging axe trap is in a larger location than the firepit trap. We already know that the firepit trap is in a location 30 square stinkomiles large, so the swinging axe trap must be in a location 40 (or more) square stinkomiles large. The only location 40 square stinkomiles large is the Fire Island Acid Pools. This means that the swinging axe trap must be in the Fire Island Acid Pools.

10: Back in step 1, we worked out that the boulder trap is on the east side of Wonderland. The only remaining location on the east side of Wonderland is the Fire Island Jungle, so that must be where the boulder trap is hidden.

11: Back in step 8, we worked out that the firepit trap is in either the Windy Hills or the Fire Island Jungle. We now know that the boulder trap is in the Fire Island Jungle, so the firepit trap has to be in the Windy Hills.

12: From clues 6 and 7, we know that the pitfall trap is in a smaller location than the treasure, but in a larger location than the spikeyball trap. The last three remaining locations have the sizes 10 square stinkomiles, 20 square stinkomiles, and 30 square stinkomiles, so it's easy to deduce what goes where.

FINAL ANSWER:
The exploding barrel trap is on Sundog Island
The fireflower trap is on Foggy Peak
The cage trap is at Crystal Lake
The chomper trap is on Lonely Top
The swinging axe trap is in the Fire Island Acid Pools
The boulder trap is in the Fire Island Jungle
The firepit trap is in the Windy Hills
The spikeyball trap is in Mushroom Grove
The pitfall trap is in Stinky's Cove
The treasure is in Wondertown
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VirtLands
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Scritterbrained

Post by VirtLands » Thu Jun 20, 2013 9:34 pm

...couldn't figure that one out....
Last edited by VirtLands on Sat Aug 10, 2013 7:45 pm, edited 1 time in total.
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Post by Yzfm » Fri Jun 21, 2013 12:13 am

@Scritterbrain: Wow, that WAS ingenious. I only managed to work out that the exploding barrel trap was in Sundog and step 2, before I gave up. West and east were confusing me and I ended up mixing them. That puzzle is fun to solve, if not mind aching. If anyone has more of those logic puzzles, share them here please. :)
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Post by Emerald141 » Fri Jun 21, 2013 1:15 am

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Post by boh123321 » Fri Jun 21, 2013 3:34 am

Emerald141 wrote:I don't check this topic often, so I only noticed the Scritterbrain puzzle after you posted the solution for it, at which point I quickly worked it out myself without the help of the solution. I FEEL PROUD.
Wow.

You are good
no
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Post by yot yot5 » Fri Jun 21, 2013 8:24 am

Emerald141 wrote:I don't check this topic often, so I only noticed the Scritterbrain puzzle after you posted the solution for it, at which point I quickly worked it out myself without the help of the solution. I FEEL PROUD.
Well done! I'm glad to see somebody managed to solve it. :lol:
Yzfm wrote:@Scritterbrain: Wow, that WAS ingenious. I only managed to work out that the exploding barrel trap was in Sundog and step 2, before I gave up. West and east were confusing me and I ended up mixing them. That puzzle is fun to solve, if not mind aching. If anyone has more of those logic puzzles, share them here please.
I'll start Wonderland-ifying one straight away!
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Post by Emerald141 » Fri Jun 21, 2013 7:30 pm

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Post by yot yot5 » Sat Jun 22, 2013 8:24 am

Emerald141 wrote:Also a bit of a glitch in the Scritterbrain puzzle:
The official answer has the boulder trap in the Jungle, and the Chomper trap in the Lonely Top. Having them the other way around doesn't contradict any of the clues, so there are really two possible solutions.
Anti-loophole wrote:No. We know from clue 9 (step 5 in the solution) that the chomper trap is on Wonderland's mainland.
And here's another puzzle for you!

THE FOURTH WALL
This puzzle was originally found in the same Usborne Superpuzzles book as before, named "Logic Puzzles". I do not own the design of the puzzle, but I own the text and Wonderlandic rewrite. This puzzle isn't quite as hard as Captain Scritterbrain's Treasure, but it's still very tricky!

The Story So Far...
You are Sir Eembi Lumba, the famous explorer, and you have just discovered a small cave on Willow Rock. On the wall of the cave, there is a plaque showing ten stinkers. Above the plaque, there is a message carved onto the wall. According to this message, the plaque shows ten famous stinkers who live beyond the Fourth Wall. Each of these ten stinkers is clever beyond belief, but one shines above the others: the legendary Midnight Synergy! You know that the museum sponsoring you has been searching for a picture of Midnight Synergy, so you decide to take the plaque off the wall and take it to the museum. Sadly, you find out pretty quickly that the plaque is MUCH too heavy.

It's then that you have a brainwave: If you used your rock-cutting equipment, you could cut the image of Midnight Synergy out of the plaque and carry that home instead! There's only one little problem: the message above the plaque didn't actually tell you which stinker Midnight Synergy was. Instead, it gave you a long list of clues. If you piece all the information together, you might be able to work out which stinker on the plaque is Midnight Synergy.

The 10th Anniversary Party
This plaque shows ten stinkers who live beyond the Fourth Wall, celebrating the 10th Anniversary of something truly extraordinary. The stinkers are:

-----------------------------------

Midnight Synergy
MyNameIsKooky
Emerald141
Yot Yot5
Wonderman109
Nobody
Master Wonder Mage
TheThaumaturge
Virtlands
Muzozavr

-----------------------------------

In the one of the pavilions, you can see five of these famous stinkers eating a meal together. At the back of this pavilion, three of the stinkers sit next to each other. At the front of this pavilion, the two remaining stinkers sit next to each other. In the other pavilion, you can see the remaining five stinkers playing games together. At the back of this pavilion, two of the stinkers sit next to each other. At the front of this pavilion, the three remaining stinkers sit next to each other. On the sketch below, you can see the ten possible seats. Each seat has a number marked on it.

Eating Pavilion
----1---9---10------
-------3---4----------

Games Pavilion
-------6---5----------
----2---7---8-------

1: Virtlands sitting in the front row of one of the pavilions.
2: MyNameIsKooky is in the eating pavilion.
3: Master Wonder Mage is in the games pavilion.
4: TheThaumaturge is sitting next to Muzozavr.
5: Emerald141 is sitting next to MyNameIsKooky.
6: Yot Yot5 is sitting at Wonderman109's left side.
7: Midnight Synergy and Nobody and in the same pavilion, but are not sitting next to one another.
8: If Master Wonder Mage is sitting in a seat marked with an even number, then MyNameIsKooky is too.
9: If TheThaumaturge is sitting in a seat marked with a number from 5-10 inclusive, then Midnight Synergy is too.
10: Midnight Synergy is sitting in a seat marked with an even number, but only if Yot Yot5 is sitting in a seat marked with a number from 5-10 inclusive.
Last edited by yot yot5 on Sat Jun 22, 2013 11:49 am, edited 1 time in total.
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Post by Muzozavr » Sat Jun 22, 2013 10:58 am

Muzuzavr
Muzozavr :wink:
Rest in peace, Kym. I hardly knew ya.
Rest in peace, Marinus. A bright star, you were ahead of me on my own tracks of thought. I miss you.
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Post by yot yot5 » Sat Jun 22, 2013 11:49 am

Muzozavr wrote:
Muzuzavr
Muzozavr :wink:
Whoops. :oops:
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Post by Emerald141 » Sat Jun 22, 2013 4:31 pm

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Post by boh123321 » Sat Jun 22, 2013 5:48 pm

I think I got close, but I got stuck :?
no
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Post by yot yot5 » Sat Jun 22, 2013 6:24 pm

Emerald141 wrote:I actually found this new puzzle of yours slightly harder than the Scritterbrain one (since it couldn't be solved as easily with a simple logic grid), but still manageable. I had fun solving it, too. :D
Solution to the Fourth Wall wrote:Here I assume that the pavilion rows face each other, meaning that 5 is left of 6. If I'm wrong and 6 is actually left of 5, just swap Wonderman's and yot yot's places; it won't affect any of the others.

Midnight Synergy: 4
MyNameIsKooky: 10
Emerald141: 9
Yot Yot5: 5
Wonderman109: 6
Nobody: 1
Master Wonder Mage: 8
TheThaumaturge: 2
Virtlands: 3
Muzozavr: 7
That's correct! You're really good at these. :)

And yes, you got me and Wonderman in the correct seats. I probably should have been clearer about which direction you're looking from.
Emerald141
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Post by Emerald141 » Sun Jun 23, 2013 10:24 pm

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yot yot5
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Post by yot yot5 » Mon Jul 29, 2013 11:43 am

FIVE TRAVELERS
I suddenly realized another loophole in this puzzle, and decided to remove it. SORRY.
Last edited by yot yot5 on Wed Aug 21, 2013 8:29 pm, edited 4 times in total.
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yot yot5
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Post by yot yot5 » Sun Aug 04, 2013 11:49 am

REMOVED
Last edited by yot yot5 on Wed Aug 21, 2013 8:29 pm, edited 1 time in total.
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Krishiv738
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Post by Krishiv738 » Mon Aug 05, 2013 9:51 am

I also have a puzzle which is similar to yot yot,s puzzle

I found this puzzle in a comic named tinkle which is only sold in India.

There are 2 dwarves,blue and red,one always tells lies and another always tells the truth,nobody knows who's who.who has the key?

Question to blue dwarf
You tell:Do you have a key?
Blue dwarf tells:yes


Question to red dwarf
You tell:Do you have a key?
Red dwarf tells:yes
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VirtLands
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... 2 dwarves, blue and red

Post by VirtLands » Sat Aug 10, 2013 6:52 pm

Considering Krishiv's puzzle,

If you ask both the red dwarf and the blue dwarf which has the key,
then it's impossible to tell, because either one can be lying and either one can be telling the truth. Image

So, without more clues, it's impossible to know, ...but I could be wrong.
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VirtLands
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some 8 puzzle questions I found

Post by VirtLands » Sat Aug 10, 2013 7:43 pm


This is a quiz of 8 questions I found. ( I did not invent these. )
So, let's see who can answer them. If you get confounded, then do feel free to help eachother.

Q-1: Johnny's mother had three children. The 1st child was named April. The 2nd child was named May.
WHAT was the 3rd child's name?

Q-2: A clerk at a butcher shop stands 5'10" tall and wears size 13 sneakers.
WHAT does he weigh?

Q-3: Before Mr. Everest was discovered, what was the highest mountain in the world?

Q-4: How much dirt is there in a hole that measures 2 feet X 3 feet X 4 feet ?

Q-5: What word in the English language is always spelled incorrectly?

Q-6: If you were running a race and you passed the person in 2nd place,
WHAT place would you be in now?

Q-7: Which is correct of the following? Image
Image (A): "The yolk of the egg is white"
Image (B): "The yolk of the egg are white"

Q-8: A famer has five haystacks in one field and four haystacks in another.
HOW many haystacks would he have if he combined them all in one field?
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yot yot5
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Re: some 8 puzzle questions I found

Post by yot yot5 » Sat Aug 10, 2013 8:10 pm

ANSWERS wrote:1: Johnny
2: Meat
3: It was still Mt. Everest
4: None
5: Incorrectly
6: 2nd
7: Neither
8: One big haystack
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boh123321
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Post by boh123321 » Sat Aug 10, 2013 8:13 pm

Q.1.
Johnny
Q.2.
The butcher weighs meat.
Q.3.
Mt.Everest
Q.4.
None. There is no dirt in a hole.
Q.5.
Incorrectly
Q.6.
2nd? I don't know for sure.
Q.7.
Neither. Yolk is yellow, not white
Q.8.
One large haystack
That was fun!
no
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Master Wonder Mage
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Post by Master Wonder Mage » Sat Aug 10, 2013 10:26 pm

Krishiv738 wrote:I also have a puzzle which is similar to yot yot,s puzzle

I found this puzzle in a comic named tinkle which is only sold in India.

There are 2 dwarves,blue and red,one always tells lies and another always tells the truth,nobody knows who's who.who has the key?

Question to blue dwarf
You tell:Do you have a key?
Blue dwarf tells:yes


Question to red dwarf
You tell:Do you have a key?
Red dwarf tells:yes
Since they both respond the same thing, it's impossible to know which one without asking at least one additional question.
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VirtLands
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yup

Post by VirtLands » Sat Aug 10, 2013 11:33 pm

Master Wonder Mage, boh123321, and yot yot5 were all correct.

Next time, there should be some puzzle that is harder.
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