Capt Ayhab

Capt'ne jan

ghorbanat, the 2$ is within the 27$ the innkeeper took not besides.
He took 27$ and gave back the guys 3 dollars that they shared 1$ each (I hope so, if Mazloom were with them perhaps he would have took the 3$ and asked for the rest).

One more step ...

Majid, you're fulfilling Irandokht's prophecy that you'd reject the correct answer. The 1st case she argues is quite correct, but she's letting others finish it. ID divided the 12 into s1, s2 & s3 w/ 4 balls in each. Compare s1 & s2, if equal ID has eliminated that case. So say s1 is lighter than s2. Here's another step that is a giant one for this puzzle, but a small one for puzzle-kind. On the scale, place 2 balls from s1 & 1 ball from s2 on one side, and 2 other balls from s1 & 1 other ball from s2 on the other side. So 2 balls from s2 are left on the side. If they balance, the odd ball is from the 2 left aside and it's heavier. Just measure one of them against a normal one to pick the odd one. But what if this last scale measurement (2 from s1 & 1 form s2 on each side) didn't balance out? Who's gonna put this puzzle out of its misery?

update

there are 2 correct answers to my HIJACKED[Thanks to agho majid] riddle, and I ain't saying which one ;-)

-YT

WooHoo.....NooHoo ;-)

That was not the question here is why:

They paid 3 * (10-1) = $27

Their $27 was spent on;

Room: $25
Tip: $2

27 + 2 = 29 - 30 = -1

kam omad ye dollar ;-)

-YT

Agho Majid

Ma mokhlese harchi hijacker ba safa mesle shoma.

-YT

Am I Too Late for the Game?

If I am not, may I try? I haven’t had time to read all the answers yet. If no one has come up with the correct answer, here is mine:

$25 to the hotel
$2 to the bellhop
$3 to the guests

The total is $30.00. The math in the riddle is misleading naturally. It is the whole purpose behind a riddle.

jaleho

you're right.  i took the path of least resistance...lol.

ok.  it breaks down into plain logistics.  many variables but it's NOT finite.  i just need more paper...lol.

babe... you better be preparing dinner while you're at "happy hour".  :-0

Majid, the blog hijacker!

ID....You're getting a leeeetle bit close, but not quiet!

Captain.......see  how easy it was to hijack your blog!.....LOLOL

Chaakerim dadash.

AF

you're right about the first case if they do balance out, your abnormal ball is in the remaining 4.

You then weigh three of the remaining 4 balls against three of the normal balls that you've put aside. 

- If it's balanced, you know the odd ball is the one you didn't weigh yet. Weigh the odd ball against one of the "normal" ones to know if it's lighter or heavier.

- If they don't balance:  You now know that the odd ball is in the three that you just weighed and you know whether it is lighter or heavier compared to the normal ones. So you take two of the three balls from the side that is either lighter or heavier than the normal side and weigh them against each other. If they balance the odd ball is the one you didn't weigh. You already know if your odd ball is heavier or lighter than normal, so if they don't balance, the odd ball is the one that follows the same logic (lighter or heavier).

I guess it could be explained more clrealy than the job I did... but you get the gist.

Now if the two groups of 4 you picked at first don't balance out, one group is heavier, one group is lighter and we already know the other group of 4 that we put aside is normal.

This is going to take some time and some typing, so I'll elaborate later if noone else answers. 

IRANdokht

As assistant to Agha Majid, fish

think again! You didn't even finish the part you thought you've covered correctly :-) :-) So, go back to your previous folder and DON't assume the the three in your second trial put together with three normal ones are equal!! In each case, you have to completely cover 3 possibilities:

1. If the two sides are equal...then next step

2. If left side is heavier and right lighter....then next step

3. If left is lighter and right is heavier...then next step

Of course you can always argue that with the symmetry, certain cases are eliminated with a similar logic.

The way to approach it is the issue Captain!

This has become like the Iranian nuclear programme, it is the question of how to look at it!

The bill came to $25,
The man kept $ 2 for himself
and gave $ 3 to the guest ($1x3)
-----
$30

All material accounted for!

They paid 3 * (10-1) =

They paid 3 * (10-1) = $27

Their $27 was spent on;

Room: $25
Tip: $2

JD, doost e aziz, NOPE!

But you got the first step right, Elimination process !!!

We have to eliminate 11 in order to find that ONE son of a gun!............LOL

And ladies and gents.

When you group the balls, call them like 3a, 3b, 3c.....or 4a, 4b......or 5 or 6 a,b,  it makes it easier to follow your logic.

I'm not going to make a comment about each and every entry, it'll take all my weekend while I myself am trying to find out how to single out that SOB!!! 

I started my happy hour already! so I can see 24 balls as we speak!!.......LOL

So........Convince yourself first that you got the answer.

I'll keep an eye on comments.  

As agha Majid's assistant,

JD, you're wrong :-)

 Assume that in the firts measurement, your two group of 3 in scale are equal. Then in the second trial, when you put three normal, and three containing the odd one, and if you get them equal again, then you have three left containing the odd one. But, you don't know if the odd one is the lighter or heavier still.

While I'm tapping my foot for the answer

The way it goes......I may extend the window to (2) months!

There's only ONE correct answer ladies and gents.

How ever you want to start and proceed, (3+3+3+3.....6+6......4+4+4......) don't forget the "lighter OR heavier" fact.

Capt. Since you think you have the answer, you can e-mail me the answer if you prefer

And.....one more hint,

Believe me, it takes more than one sheet of paper to write down all possibilities

JD

It seems logic. I had a hint like this for beginning with, but got lost in the way!! I think Majid and Jaleho were right, I should write it down.

Let see what the agha moalem will say about your answer.

ID... good so far

ok.  i weight 4 against 4  (1st weight)

if they are equal, you know the 3rd batch of 4 contains the unequal one.  so you would weight 3 from either the 1st set or the 2nd set (because you know they are equal) against 3 from the 3rd set. (2nd weigh)

if they are equal, then you know it's one of the remaining 2 (one from the good set and the remaining one from the 3rd set.

the 3rd weigh would be between those 2 to determine light or heavy.

BUT... what if the first 2 sets of 4 are NOT equal.

damn.  back to the drawing board.

Majid's math queeze

Dear Majid
Elimination process:

divided the twelve balls into 4 group of three
On Scale:
3s1
3s2

Out of scale:
3a
3b

STEP ONE:
tells us which group has a set of 3 with an odd. If scale is in equilibrium then the set of three is out of scale. If scale is not in equilibrium then set of three with an odd is on the scale.

STEP TWO:
From the group take one of the set of three from the group that has an odd in it and put it on one side of scale and take a set of three from the side that did not contain and odd one and put it on the other plate of the scale. In this step we simply determine that the odd one is either heavier or lighter. If the scale is in equilibrium, we know the odd one is in the set of three with odd that is left outside, otherwise it is the set of three in the scale which is different from the other plate's set. We now know which set of three has an odd and is it lighter or heavier, step two is ended.

STEP THREE:
Take the set of three with an odd in it and the fact that you know one of these three is odd and you know if it should be lighter or heavier. In this step you put one on each side of scale and one outside of scale. The side that is heavier or lighter based on known fact of step two has the odd ball and if they are in equilibrium the odd one is sitting outside of scale.

I am sure there are easier way to explain it mathematically. But is this correct?

ID/AF

ID: I was thinking about the same way. We shouldn't wight all of them  together at the beginning. We must eliminate first a group of 6, or 4, or even 2 ??? I don't know about the rest :-(

AF: I agreel with you, dear. NO sugar for Majid tonight :O))

real hint would be...

"start to weigh in 4 against 4 and leave 4 aside..."

migam haa

nakoneh shoma ham mesleh Captain javab hayeh dorost ro ham ghabool nakoni

IRANdokht

HINT

I gave you guys what? ONE MONTH? for what reason?

NOW, go ahead and do some homework........LOL

Khoob gozaashtametoon sar e kaar ha!......;-)

Find the Mehdi by juggling the balls!

In Majid's puzzle you don't know if the odd ball is lighter or heavier, a priori. In fact, Majid was too generous in his expectation. Not only should you be able to determine which of the 12 is the Mehdi, I mean the odd ball, in 3 weighings, but also whether the odd one is lighter or heavier. It involves several steps to go through all the possible outcomes.

souri

i'm wrong so it doesn't make any difference but that last (3rd weighing example would be "either / or"...

it's only 3 weighs... :-)

dang it.  english major / math minor.  can't spell and can't solve a riddle...lol

let me get another binder.  i've filled this one up.  damn it Majid.  no sugar for YOU tonight.  :-)

Please pay attention

One out of 12 is either heavier  OR  lighter ! you don't know!

The catch is....you don't know how different that ONE is!

I promiss you, it's NOT as easy as you think it is!

OK..... 6 and 6, not ballance of course, which 6 are you going to eliminate?

4 and for, not equal! which 4 you're going to focus on?

Or.....4 and 4, equal! now you used the scale once and now you have 4 with one different.  and 2 try remaining 

Fish and Souri, right

Here's Majid's emphasis:

"ONE out of 12 is different in weight, "either lighter OR heavier" 

AF.....

You've got 4 weighting .......we can weight them only 3 times

jaleho

my error is assuming that we're looking for a "lighter" or "heavier" ball specifically, right? 

Jaleho

I know there should be a mistake somewhere, otherwise Majid jan wouldn't give that so easy riddle to solve here. but honestly, I didn't get  your question.

In my answer, I take the case that "one of the 12 ball is heavier than the rest"........

Now, regarding your answer, I believe I got it wrong. You mean, on of the ball is either heaviest or lightest, but we don't yet know what is the case? Did I get your point?

I thought we have two distinct cases here.

ok... let me think

1st weighing... 6 balls versus 6 balls

2nd weighing... two groups of 2 from the 6 that were lighter

3rd weighing... if 2 groups of 2 are equal, weight the remaining 2 to determine which one is different.

same thing if the 2 groups are different, the 3rd weighing would be the remaining 2.

that's too easy. what's the catch?

Souri, here's your mistake

"2) Now, divide the 6 balls of the heavier kaffeh in two groups of 3 balls, each in one kaffeh. Put aside the balls in lighter kaffeh "

What if the next time around the two sides came equal. That is, the 6 balls that you set aside in the first trial, contained the lighter ball? Can you figure out the lighter ball out of six in one trial?