The Brilliant 9 Coin Riddle! Best Riddle Ever!

[forgingiron123]

Step One: Three coins on each side.

Step Two: If they balanced, weigh two of the three previously unselected coins, or if they didn't balance, the heavier side. If they balance, the last coin is the heavy one. If not, it's (obviously) the heavier one.

[misteriddle]

@forgingiron123 Brilliantly explained! =D

[xShadow125]

Uhm, put one coin in each side of the balance, one by one, one goes in left, and one in right, then keep doing that until you get a coin that makes it uneven, if you get to 8 (four right and four left) and it still not uneven then the last one left should be it.

[misteriddle]

@xShadow125 The thing is, it has to be done in 2 moves, but if you put them in one by one, it could take up to 8 moves...

[MrPoohbear2218]

you can do it with four coins eack as well put four on each side if it leans the the heavier side has the coin if it doesnt lean the coin left out is the coin you are looking for

[misteriddle]

@MrPoohbear2218 Let's say it does lean to one side, how would you know which coin it is from that pile of 4 coins?

[TheBbk8]

Weigh any 3 on one side and any 3 on the other. If you find both weigh the same, then the heavier coin is in the remaining 3. In this case, you take any 2 of those and weigh them 1 per side. You will see that the heavier coin is being weighed, or the 2 coins are the same weight so the heavier coin remains. In the instance that after the initial weighing you find one side heavier than the other, repeat the process of weighing the 3 again by means of 1-to-1 and finding the heavier as before.

[misteriddle]

@TheBbk8 Excellent explanation!

[Dimcair]

took me about 4 min

[misteriddle]

@Dimcair You're pretty clever!

[blendistars]

Here's a riddle for you misteriddle i have four coins in my left hand, i take two away, how many coins do i have?

[misteriddle]

@blendistars Hmmm.... 4?

[blendistars]

@misteriddle no 8 because i also had four in my right hand :P

[misteriddle]

@blendistars Erm... I think I get it!

[iceman04able]

huh?

[misteriddle]

@iceman04able Erm...

[547dagger547]

Very interesting and quite fun to think about! However, after a few minutes, it becomes pretty obvious :(

[misteriddle]

@547dagger547 Good to read that you worked it out!

[joda294]

man is your name cvetan?

[misteriddle]

@joda294 Erm, Nope... Why do you ask?

[argumentativeindian]

I initially thought it'd be 4 against 4, but that could take 3 weighings. 3 against 3 would get it down to exactly 2 weighings, like MultiMitchell7 explained. Hey, @misteriddle do you have the version of this puzzle, where there are 12 such coins, and we know that exactly one from amongst these 12 is different, but do not know whether it's heavier or lighter than the rest, and we need to find that coin in a max of 3 weighings..?

[misteriddle]

@argumentativeindian Yep, I've heard of that riddle. It has a longer explanation though, doesn't it? If I have time I'll try and make that video as well... Bit busy at the moment. ;p

[MultiMitchell7]

SPOILER ALERT!!!

you can weigh 3 coins on one side and 3 coins on the other.

If neither side leans, you weigh 2 of the coins that you left out at first. If the scale doesn't lean, then the coin is the one you left out.

If the scale leans at first, you put 2 of coins of the heavier side on. If the scale doesn't lean then, the coin is the one that you left out

[misteriddle]

@MultiMitchell7 Very nicely explained!