TimeSpaceLightForce

@Delta 1212 You're on the right track! The given weight of normal coin is X ..usually varies from 2.1 - 8 grams , but for the purpose of the puzzle assume 1 gram. ( I edited this OP w/ "1 unit like the other .." but instead had double posting .. and that was deleted )

@Imatfaal It is easier than the classic 12 balls with odd one to find by balance scale in 3 trials.. @John Cuthber Yes, It can be spoiled later.

The coins shall be weighed by digital weighing scale not the balance scale..

One of the 12 identical coins was magnetized so that it weighs (Z or Y) heavier on one face and lighter on the other face when place on a certain weighing scale. But it weighs X like the other normal coins when it is standing balanced on its side or edge. Using the weighing scale just 3 times ..find that odd coin and determine its heavy or light face.

I was directed to almost similar arrangement of this called Swinging Atwood Machine ( also known as smiles & tears). Its a lot complicated than i thought. Below the red circles are pulleys. O -- -- -- --O-- -- --o | | | | | | o

thanks.. but the imagine the lower ball moving downward if there is a horizontal tug on the swinging ball. Maybe your experiment has high friction..

J -- -- -- --o | | | | | | o Attached on a 1 unit long and weightless cord are 2 identical balls w/ 1 unit mass. If it slides on hook J w/out friction.. the high ball can be pulled horizontally to distance (d) away from J so that when released it can hit the other ball. Can I assume that the low ball is at rest on time of collision?

What would be the thickness of the coin (disk or cylindrical) with 1 unit diameter that would make the tail /head/body(side) outcome be a fair 1/3 probability?

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

This is the complete version: Multiply Values z e r o = 0 a = ? o n e = 1 b = ? t w o = 2 d = ? t h r e e = 3 e = ? f o u r = 4 f = ? f i v e = 5 g = ? s i x = 6 h = ? s e v e n = 7 i = ? e i g h t = 8 l = ? n i n e = 9 m = ? t e n = 10 n = ? t w e l v e = 12 o = ? t w e n t y = 20 r = ? f o r t y = 40 s = ? s i x t y = 60 t = ? s e v e n t y = 70 u = ? n i n e t y = 90 v = ? 3 h u n d r e d = 300 w = ? 5 t h o u s a n d = 5000 x = ? 8 m i l l i o n = 8000000 y = ? 1 b i l l i o n = 1000000000 z = ? Use excel. It's like playing Sudoku..

hey guys, this one is new and not easy! Get the values of factors or variables : l e f t v o x w i n s u g h r that can satisfy all equations below ... o n e = 1 t w o = 2 t h r e e = 3 f o u r = 4 f i v e = 5 s i x = 6 s e v e n = 7 e i g h t = 8 n i n e = 9 t e n = 10 e l e v e n = 6 t w e l v e = 12 note: You could add subtract multiply or devide the words with numbers so that twelve + one =13 = 4two+nine/3+7-five eleven + two = twelve+ten+8/two+7nine/seven+one-six5+4/2 so on.. have fun.. pls use spoiler

basic algebra: if a=b and c=d then ac = bd eq:I or ad = bc eq:II as da = cb also a + c = b + d eq:III or a + d = b + c using the above derive the anagrammatist equation from below : 1. t = o n 2. e = o w 3. l = w t 4 . e2 = n v 5. e3 = v l

Yes, im sorry.. but you can try again or find the solution posted on this thread..

That is a superb path.. though still has one switch off.

Alright ! You got it Right.. .. I fail to state that- the Laser Switch center needs enough energy continuously (i.e. about the gun's output) thus making: 4 one-way mirrors fail due to decreasing energy 3 mirrors fails to hit the centers 1 rotating mirror fails to supply continuous energy 5 mirrors fails for too much energy on some switches 1 spherical mirror fails.. too low energy 0 mirror fails, near zero energy So we can still try solving the original puzzle.. Thanks Rouie

Other solutions . .

Beam along (1,3 ) hitting switch at (1,4) with ( \ ) 45deg mirror inside circular field switch will pass thru (2,4) switch.. turning both on.

There are 16 [off]switches that can be activated by passing a laser beam into them.Using 5 mirrors and a laser gun , how can you switch them all [on] with one shot? O O O O O O O O O O O O O O O O Please use