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# Weighing in a Harder Way

Best Answer bonanova, 18 January 2008 - 06:49 PM

Jkyle1980's solution is correct, and much less verbose than what follows.

You can distinguish among 3**N cases in N weighings.
There are 54 possible cases in this puzzle [one of 27 coins is heavy or light].
So three weighings [27 cases] won't do it, but four [81 cases] can.

Spoiler for Here's a very slightly modified approach
Go to the full post

83 replies to this topic

### #51 Shachi

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Posted 23 September 2009 - 01:12 AM

I came up with what, to me, seemed like an even simpler explanation.
You have 27 coins...
Divide them into two groups of 13, holding one coin out.
the measurements on the scale will read 130g for one group and either 130, 131, or 129 for the second. Disregard the group with the weight of 130g.

Assuming you didn't pull the odd coin, again split the coins in the lighter/heavier group in half once again leaving out one coin
This time the scales would read 60g for one group and 60/61/59 for the other

The third time, you'll have no need to remove a coin, so just weigh the two groups of three.
Same as before the scales will read 30g for one and either 30/31/29 for the third.

You're now down to three coins. Hold one out and weigh the final two, and voila!

Edited by Shachi, 23 September 2009 - 01:13 AM.

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### #52 ogden_tbsa

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Posted 23 September 2009 - 08:22 PM

I came up with what, to me, seemed like an even simpler explanation.
You have 27 coins...
Divide them into two groups of 13, holding one coin out.
the measurements on the scale will read 130g for one group and either 130, 131, or 129 for the second. Disregard the group with the weight of 130g.

Assuming you didn't pull the odd coin, again split the coins in the lighter/heavier group in half once again leaving out one coin
This time the scales would read 60g for one group and 60/61/59 for the other

The third time, you'll have no need to remove a coin, so just weigh the two groups of three.
Same as before the scales will read 30g for one and either 30/31/29 for the third.

You're now down to three coins. Hold one out and weigh the final two, and voila!

this would work great if you had a 3 pan balance, and used the 3rd pan for a known weight, e.g. 130g.
Without having a weight reading is the real challenge.
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### #53 kyky

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Posted 02 October 2009 - 02:50 AM

my answer is 5 times.. heres how it goes

27 coins = 8 8 8 1 1 1. thats a b c d e f
if a = b, a = c, we go to the ones.. if d = e, then the answer is F. COUNT = 3
if a = b, a = c, we go to the ones.. if d =/ e, we weight f. if d =/ f then the answer is D. COUNT = 4
if a = b, a = c, d =/ e, d = f, the answer is E. COUNT = 4

now we go to the if the diff coin is at a or b or c.
if a = b, a =/ c, then we measure c.

before we go further, i want to show you how we solve 8 coins with 3 scaling.
8 coins we devide it into twos.

8 coins = 2 2 2 2, thats w x y z.
if w = x, w = y, then we weight 1 coin from w, and 1 coin from z. if w = z, then the other coin from z is the answer.
if w = x, w =/ y, then we weight 1 coin from w, and 1 coin from y. if w =/ y, the answer is that coin from y.
if w =/ x, w = y, then we weight 1 coin from w, and 1 coin from x. if w = x, the other coin from x is the asnwer.
if w =/ x, w =/ y, one coin from W is the answer. we weight 1 coin from W, 1 coin from x or y or z..
TOTAL COUNT = 3

if a = b, then the coin is at C. total count 4.
if a =/ b, a =/ c, then the coin is at A. total count 5.
if a =/ b, a = c, then the coin is at B. total count 5.

Edited by kyky, 02 October 2009 - 02:52 AM.

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### #54 kyky

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Posted 02 October 2009 - 02:55 AM

You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g).
The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter?
I have no answer to this question.
Do you?

look at my answer =) hope helps.
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### #55 Nikk29

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Posted 02 October 2009 - 11:38 PM

Yes... I seem to find this problem a tricky one... I heard it another way (the one coin is heavier). If the one coin was heavier, then the answer would have been
Spoiler for ---

But because your coin is either heavier or lighter and you dont know, you have to check it
Spoiler for ---

I love this one!

Spoiler for ---

I know I am kind of late on answering this... but I just wanted to go back to it. FACE PALM!

JUST CLARIFICATION:
Spoiler for ---

I hope this post helped at all.

Edited by Nikk29, 02 October 2009 - 11:42 PM.

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### #56 Nikk29

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Posted 02 October 2009 - 11:45 PM

look at my answer =) hope helps.

Note, this problem has been solved earlier in the postings. The answer is not what you got, but you did great answering it. You can read another person's post, and hopefully it helps you understand. You DID do well in explaining your answer though.
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### #57 kyky

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Posted 04 October 2009 - 04:04 AM

Note, this problem has been solved earlier in the postings. The answer is not what you got, but you did great answering it. You can read another person's post, and hopefully it helps you understand. You DID do well in explaining your answer though.

thanks for the compliment, but no i dont read others answer before i got mine.
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### #58 varin

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Posted 27 October 2009 - 02:26 PM

5

Edited by varin, 27 October 2009 - 02:31 PM.

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### #59 varin

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Posted 27 October 2009 - 02:30 PM

Spoiler for answer is 5

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### #60 satanite

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Posted 14 January 2010 - 02:11 PM

In minimum 6 attempts one can get the definite answer.
Make 27 as 9,9,9 we say (pile1,pile2,pile3)

1.Weigh 2 piles (pile 1 & pile 2), If they are balanced then the coin is definitely in 3rd pile (pile 3) .
2.If not then again weigh either of the pile(pile 1) with the remaining pile (pile 3).
(If both sides are balanced the different coin is in the pile 2, If not its in pile 1)

Now we are having 9 coins, again make 3 piles of 3 coins in each

3.Repeat step 1 with new 3 piles (pile 4,pile 5, pile 6)
4.Repeat step 2

Now we are having 3 coins,

5.Weigh 2 coins (coin 1 & coin 2), If they are balanced then the 3rd coin is different.
6.If not then again weigh either of the coin (coin 1) with the remaining coin (coin 3).
(If both sides are balanced the different coin is coin 2, If not its coin 1)

Edited by satanite, 14 January 2010 - 02:12 PM.

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