Humitidy +1

Answer:

Ten. Every cell that is a perfect square will remain open
(1, 4, 9, 16, 25, 36, 49, 64, 81, and 100).  If a number is not
a perfect square, then it has an even number of divisors,
therefore it will be "toggled" an even number of times and
end up where it started (closed).  Perfect squares have an odd
number of divisors, so they will end up the opposite of where
they started (open).

Weigh 3 coins on each side...take those coins off and weigh the remaining set of 3 coins (so 6 coins in total, 3 on each side).

From here, you will know which set of 3 coins contain the "odd coin" because it will either be heavier on one side or lighter on one side of the above weighings.

EDIT: Ok, I will clarify this part a bit more...if the first 3 v 3 weighing is equal, you take off only 1 side and weigh another 3, if it is different, you will know if it is lighter or heavier...you then weigh the lighter/heavier side as per below...

From there, weigh only 2 of the coins, if they balence, then the unweighed coin is the odd coin, if not, you will know which side has the odd coin...

weigh 5 and 5, if the scale is equal your 1 odd is the correct
if the 1 odd is in the mix of 5 weight 3 and 3 and if the odd is in the set of 3 weight 1 and 1 coin with 1 remanding, if the 2 on the scale = same the odd is in your hand if the scale shows a differance the odd coin in winner.

6 vs 6
whichever one weighs more split in half and measure - should either be different or equal
if equal dismiss this 6 use other 6
if different use this 6 and dismiss other 6

with your final 6 you will now know wether it weighs more or less due to first test.
split in half - if from your first test you know it weighs more whichever 3 weighs more contains the odd coin
- if from your first test it weighs less you know the 3 coins that weigh less contain the odd coin

from the final three - put one on either side..
if equal the other coin is the odd one
if different the one that weighs more is the odd one.

Reword the riddle otherwise Arby is right

Wornstrum +1

Answer:

This is a real toughie.  Along the way, we'll denote coins
we know nothing about with '?'; coins that can't be the odd
coin with '0'; coins that must be heavier if they're odd with 'H';
and coins that must be lighter if they're odd with 'L'.
STEP 1. Weigh four coins against four coins.  Either (a) they
balance or (b) they don't.  So (a) looks like 00000000???? and (b)
looks like HHHHLLLL0000.
STEP 2(a). Weigh two '?' coins against one '?' and one '0'.
There are three possible results: (i) they balance, (ii) the '??'
side is heavier, (iii) the '?0' side is heavier.
STEP 3(a)(i).  Since the scales balanced again, we now have
00000000000?.  The last coin must be the odd coin.  Weigh it against
an '0' coin to see whether it's too heavy or too light.
STEP 3(a)(ii). The 4 coins not weighed in step 1 now are HHL0.
Weigh H against H.  If the scales balance, L is the odd coin and is lighter
than the other coins.  Otherwise, the tip of the scale tells you which
is H is in fact heavier and which is actually an 0 coin.
STEP 3(a)(iii).  Proceed as in 3(a)(ii), but with L's and H's switched.
STEP 2(b).  We so far have HHHHLLLL0000.  Weigh HHL against HL0.
There are three possible results: (i) they balance, (ii) the 'HHL'
side is heavier, (iii) the 'HLO' side is heavier.
STEP 3(b)(i). The five "candidate" coins on the scale in the last
step must be ok, leaving HLL000000000.  Proceed as in step 3(a)(iii).
STEP 3(b)(ii).  The three "candidate" coins not weighed in the
last step must be ok.  Also, since HHL was heavier than HL0, either
an H from the left side is truly heavier or the L on the right
side is truly lighter.  This gives HHL000000000, so proceed as
in step 3(a)(ii).
STEP 3(b)(iii).  Again, the three "candidate" coins not weighed
in step 2(b) are ok.  Since HL0 was heavier than HHL, either
the H on the left is truly heavier or the L on the right is
truly lighter.  Weigh the H coin against any 0 coin.  If they
balance, the L is truly lighter; if not, the H is truly heavier
(and the scale will tip toward H).

yeah nice one I was just starting that road now, but you beat me to it

> ~Wornstrum~ wrote:

> Nope, it is fine, the odd coin is lighter or heavier than the other coins...

The riddle stated the answer basically

"You have twelve coins, eleven identical and one different."

So technically Arby was right! There was no mention that they all were identical

put on the light?

Take the coins, and place them into 2 piles of 50 coins each, both on their sides. That way, all coins will have 0 tails facing up, and will be equal.

Sir_Prozac +1

Wheeeeeee