[The last Airbender]

Episode 25

The Lion Turtle & the spirits
are looking for it in the 4 elements
Summed the Master to study fire
Sent a warrior to see earth
Let the bender to search water

When The Avatar find it in air..
He told the Lion Turtle that he have it.
When all the others came back
they all declared they have it too.

Now the Lion Turtle and the spirits
know they had what the 4 elements has

But who have it first?

hey there guys ,

your riddles are quite difficult but very nice .its my first post here.thanks

[X MAS TREE]

I guess this is unanswerable after all..

in most escape games point and clicking

some sign can give you the right number usually

need to open a coded box, door,compartment

that has another item inside:

X means multiply

M 11^ flip & cut

A - 4 breaking

S 8 meeting ends

T 7 - dropping

R 1 2 separating

E E 6

9 rotate and link..

this is how to do it in games:

but still don't make sense...

[unanswered]

This is a 21 unknown 21 equation problem:

a = ? zero = 0
b = ? one = 1
d = ? two = 2
e = ? three = 3
f = ? four = 4
g = ? five = 5
h = ? six = 6
i = ? seven = 7
l = ? eight = 8
m =? nine = 9
n = ? ten = 10
o = ? twelve = 12
r = ? fifteen = 15
s = ? twenty = 20
t = ? thirty = 30
u = ? forty = 40
v = ? sixty = 60
w =? hundred = 100
x = ? thousand = 1000
y = ? million = 1000000
z = ? billion = 1000000000

I hope some other puzzle can be derived from this like: " what word is posible for 18896063730041700?"

i.e. = brainden but that is difficult and i dont know if there are other posible answers

Congratulation to the team work participants that answered this!

[unanswered]

Good job k-man! I sure did accidentally lose the "w".

And, TSLF, that puzzle is a TDF (tour de force)!

I used excel for this.. just divide and multiply easy

the w is the first to give answer from twenty= w(ten)(ty)=w(10)(10)=20

thanks CaptainEd and all the solvers..

[unanswered]

What is the product for unanswered = ?
If letters are factors of equations
zero = 0
one = 1
two = 2
three = 3
four = 4
five = 5
six = 6
seven = 7
eight = 8
nine = 9
ten = 10
twelve = 12
fifteen = 15
twenty = 20
thirty = 30
forty = 40
sixty = 60
hundred = 100
thousand = 1000
million = 1000000
billion = 1000000000







note: i got the idea for this puzzle post from X-MAS TREE thread that was unanswered by dtdt..thanks

happy holidays!

[Blood runs thick in Transylvania]

+- -- ++ -+

sv iv sh ih

if M is -+ +? -! (-++-)= +

L tells +

if M is +- +? -!

L tells +

if M is -- +? -!

L tells -

if M is ++ +? -!

L tells -

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

if L is -+

L tells -

if L is +-

L tells -

if L is --

L tells +

if L is ++

L tells +

Minna is insane human leads to Lucy is insane vampire

Lucy can not be sane human if she tell true

Lucy can not be insane vampire if she tell true

If minna is insane vampire can be sane vampire but one is human

if Lucy is insane can not tell false because both are insane.

If minna is sane human Lucy can tell false but both not insane

..not sure just follow the signs

[Blood runs thick in Transylvania]

+- -- ++ -+

sv iv sh ih

if M is -+ +? -! (-++-)= +

L tells +

if M is +- +? -!

L tells +

if M is -- +? -!

L tells -

if M is ++ +? -!

L tells -

if L is -+

L tells -

if L is +-

L tells -

if L is --

L tells +

if L is ++

L tells +

[Lethal Solution]

actually the lever method is the best next to pipet solution ..

or

Take glasses 1 2 3 4 off the scale

Spill some of 2 then refill from 1

Weigh mixture 2 with 3. If reading:

a bit > A+A : glass1 has it

a bit < A+A+x :glass2

=A+A+x :glass3

=A+A :glass4

note: taking the 4 glasses at the same time with two hands needs caution

the purpose of spilling some of glass2 content is to make sufficient

difference on poison weight (e.i about 2% to 20% water volume).

The scale can read milligram anyway 2%gram=20mg 98%gram=980mg.

Warning: DO NOT try this at home !

[Lethal Solution]

on the laboratory bench next to the digital scale, place a knife edge as a fulcrum, and balance a plank of length 2 meters on the knife edge, with one end touching the digital scale.

At this point, the plank is adding exactly no weight to the scale.

Now,

* place G1 on the plank directly over the scale,

* place G2 on the plank at a distance of 1/2 from the fulcrum

* place G3 on the plank at a distance of 1/3 from the fulcrum

* place G4 on the plank at a distance of 1/4 from the fulcrum

* read the scale

* Without the poison, the scale would read G1 + G2*1/2 + G3*1/3 + G4*1/4 will be (A+A/2+A/3+A/4) =

(12A + 6A + 4A + 3A)/12 = (25A/12)

However, the poison of weight X will either appear as an addition of X (if in G1), X/2 (if in G2), X/3 (if in G3) or X/4 (if in G4).

(Firefox doesn't seem to play well with this spoiler editor.)

I can not say no to this one CaptainEd it is so resourceful and using mechanics is cool..using knife and bench and plank.

You might as well look at the corner cabinet for the poison scanner

[Lethal Solution]

I'm not sure I would accept this challange. But if I had to , place two glasses on scale. Either these two, or the other two weigh 2A+x. We now know that the poison is one of of the two heavier pairs. Now the hard part : drink one of these glasses. If you don't die, you know the other glass is poison. If you die, you go to your grave knowing you were also correct

This is one solution if you have a goldfish in the bowl

[Lethal Solution]

Badhri's principle is to construct a situation with differing fractions of the four glasses, so that we can compute which glass it was that had added mass.

However, the OP says that the glasses are "full", which means there's no room to add from one into another.

So, here's an alternative:

(1) I gather that we are given:

* A, the weight of one glass full of water

* X, the weight of added poison.

(2) remove all glasses from the scale, without peeking at the scale.

(3) Using a volume measuring device, such as a pipet, measure the volume in G1, and empty it, call that volume V.

(4) measure V/6 from G2, transfer to G1

(5) measure V/3 from G3, transfer to G1

(6) measure V/2 from G4, transfer to G1

(7) At this point, G1 is full

(8) place G1 on the scale.

(9) If G1 weighs A, original G1 had the poison

--if G1 weighs A + X/6, original G2 had the poison

--if G1 weighs A + X/3, original G3 had the poison

--if G1 weighs A + X/2, original G4 had the poison

if you have estimated half a glass and it weighs more than half the water weight, the additional weight can be water or poison's..thats why the volume must be precisely measured as done here..still that is troublesome.. the glass weight shall be considered too in that case

[Lethal Solution]

If using the LCD screen, I would put all 4 on the scale, and remove one until one took off more weight than the others. only using the scale once. Or you could go the opposite route, and you would have to put them all on the scale one at a time as to not spill. So just watch the weights as you put a new cup on and you would know which one had the poison.

Your solution is right unless making the machine do its job means using it

(in that case reading weight trice). It is resting if nothing is on it. If there is

a reading already using once means having a new reading..

Use the scale tray if you would like take them off or place them on the scale

[Lethal Solution]

I totally agree--Badhri's principle is much more practical in terms of masses.

..too many moves and uncertainties

[Lethal Solution]

I guess we are assuming that some mass has been added. So, if there's no change in volume, the vessel with the poison must have a higher density than the others. So, without even using the scale, we could perhaps measure density in each vessel, and declare the one with the highest as the poisoned one. Since we don't know what mass of poison has been added, I don't know whether this is a realistic approach or not. For example, if only one molecule of the poison has been added, we won't be able to measure the change in density.

Interesting problem, TSLF! Thanks.

Right CatpatinEd in terms of Physics..but say we have a simple chemistry problem and 1:100 x to A ratio is observed..

[Lethal Solution]

Not sure!!!!

Take away all the glassess first (since i have to use it only once).

Let me have the glassess as g1,g2,g3,g4

pour one quater the amount of g2 to g1 say g1+g2/4

same half and three fourth of 3rd and 4th respectively

so the first glass will now have g1+g2/4+g3/2+g3*3/4 (Assuming we are able to measure g2/4 , g3/2 or g3*3/4)

now weigh that glass if the weight gonna be

CASE 1 :(A+X)+A/4+A/2+3A/4 then g1 is poisoned

CASE 2: A+(A+X)/4+A/2+3A/4 then g2 is poisoned

CASE 3 : A+A/4+(A+X)/2+3A/4 then g3 is poisoned

CASE 4: A+A/4+A/2+3(A+X)/4 then g4 is poisoned

That is a good solution :(Assuming we are able to measure g2/4 , g3/2 or g3*3/4)

Let weight be B. Then (B - 2.5A)*4/x is the number of the glass ..but less practical solution

[Foiled Again]

..more for completion

8'' 4' > 1' 11 3

8'' 4' > 1 11 3'

8' 4' > 1 11 3

8' 4 > 1 11 3'

From above broken posts :

If 8 4 ? 1 11 3

is balanced pick 1 + 3

if tilted right pick 4

if tilted left pick 4 /11

This confirms Prime's solution.. thanks

[Lethal Solution]

=========================

_____||||_____

/ (_____) \

[____________._]

'' ''

Pls. use the weighing scale w/ LCD display above,

not the beam balance below

============ = ============

|| ||

"=========[ ]========="

||||'''''''''''||||

[Lethal Solution]

On top of a digital weighing scale
are four identical glasses, filled with
water, weighs A grams each. Then,
somehow the poison X was sneaked
into one of them with no change in
color, odor or volume.

\ / \ / \ / \ /
\__/ \__/ \__/ \__/
==========================
[4 A + x ]

Using the scale once.Can you find out
which of the glasses is poisoned?

[Foiled Again]

8'' 4 > 1 11 3"

missed again..

[Foiled Again]

if tilted

8 < 8 11
try lighter

none or one
4 = 4
4 < 4'

if balanced

8 = 8 11
distribute by color
8' 4' = 1 11'' 3
8' 4 < 1' 11'' 3
8' 4 < 1 11'' 3'
8 4 < 1 11'''' 3
8'' 4'' > 1 11 3

Pls. note the previous post..TY

[Foiled Again]

If Tilted
13 < 13*** 1*
13* < 13** 1*
13* < 13*** 1
13 < 13**** 1
try lighter
6 < 6 * 1
6 = 6 1
6 = 6 1*

If Balanced
13** = 13** 1
try any
5* = 5 * 3
5 < 5* 3*

Hi ,i just joined this awsome place you have ..

i like puzzle solutions , thanks

[Greetings to Everyone]

Hey there guys,

I also like to solve puzzles/riddles.

My online past time is "scape games"

with cool puzzles to solve. Exit is the

the only way. I hope to learn from you.

And contribute to this amazing forum.

Thanks

Rouie