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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
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Posted 06 February 2011 - 03:12 PM

THE ANSWER IS SIX
When the coins are divided in three groups, two measures are enough to decide which group is different.
Thus divide the coins in three groups of 9 coins and measure them twice.
Then divide the different group in three groups of 3 and measure them twice again.
Then divide the different group in three coins and measure them twice.
TOTAL = SIX
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### #72 b1soul

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Posted 16 March 2011 - 07:33 AM

THE ANSWER IS SIX
When the coins are divided in three groups, two measures are enough to decide which group is different.
Thus divide the coins in three groups of 9 coins and measure them twice.
Then divide the different group in three groups of 3 and measure them twice again.
Then divide the different group in three coins and measure them twice.
TOTAL = SIX

that was my initial answer too (3*2, measuring it twice after each time you divide by 3)

but I realized that after dividing 27 into three 9-coin groups and measuring twice, you'll know whether the odd coin out is heavier or lighter, after that you'll only need to measure once after each time you divide by 3

6 is the answer if the weighing device only tells you whether two groups of coins are the "same" or "different" in weight (no weighing device that I know of does only that)
if the weighing device tells you which group of coins is heavier and which is lighter, the answer is 4

Edited by b1soul, 16 March 2011 - 07:35 AM.

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### #73 jayeshk

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Posted 30 May 2011 - 01:21 PM

We can do it in 3 Measurements

Make 3 Groups of 9 Coins Each

Measure1 : Weigh Any Two Groups , If they weigh same . 3rd which is kept aside contains the required coin else one with heavy side has the required coin.. This Eliminates 18 Coins out of 27

Measure2: Elliminates 6 Coins Going by Above Process . Balance is now 3

Measure3: Weigh Any 2 , if same the coin kept aside is required coin else the one on heavy side
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### #74 Kia

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Posted 26 September 2011 - 06:41 PM

The answer is not 3 as suggested above because you don't know if the different coin is heavier or lighter. With 3 attempts you can only do this if there are 12 coins. This can be done in 4 attempts. Actually using 4 attempts, you can find the different coin and know if it is heavier or lighter given 39 coins (not just 27). I will try to explain:

We have 3 groups of 13 coins. Compare 2 groups of 13. (measure #1)
if even, the different one is in the 13 that was not compared.
...Take 9 of those and compare them against 9 coins that you know are the same (measure #2)
...if even, then the different one is one of the remain 4.
......Take 3 of those and compare them against 3 coins that you know are the same (measure #3)
......if the same, then the different one is the one not measured
.........Compare the last coin against any other coin and see if it heavier or lighter (measure #4)
......if not the same, then it is one of the 3, and we know if it is heavier or lighter
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know if it is heavier or lighter
.........if not the same, since we already know the different one is heavier or lighter, then that is the one
...if not even, then it is one of the 9, and we know if it is heavier or lighter
......divide these 9 into three groups of three and compare 2 of the groups (measure #3)
......if the same, then it is one of the three that we didn't measure
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know if it is heavier or lighter
.........if not the same, since we already know the different one is heavier or lighter, then that is the one
......if not the same, then since we already know that the different one is heavier or lighter,
......then we know which group of the 3 it belongs to
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know if it is heavier or lighter
.........if not the same, since we already know the different one is heavier or lighter, then that is the one
if not even, then the 13 the we put aside are the same, and so then take out 9 of the coins from the one side, say the heavier side, move 9 coins from the lighter side to the heavier side, and add 9 of the coins from the pile that we didn't measure to the lighter side, and measure again (measure #2)
...if the same, then it is one of the 9 coins that we took out, and we know the different one is heavier
......divide these 9 into three groups of three and compare 2 of the groups (measure #3)
......if the same, then it is one of the three that we didn't measure
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know it is heavier
.........if not the same, then it is the heavier one
......if not the same, then we know it is on the heavier side
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know it is heavier
.........if not the same, then it is the heavier one
...if the scale changed direction, then it is one of the 9 coins that we moved from the lighter side,
...and we know the different one is lighter
......divide these 9 into three groups of three and compare 2 of the groups (measure #3)
......if the same, then it is one of the three that we didn't measure
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know it is lighter
.........if not the same, then it is the lighter one
......if not the same, then we know it is on the lighter side
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know it is lighter
.........if not the same, then it is the lighter one
...if the scale remained the same, then it is one of the 8 coins (4 on each side) that we didn't move
......take out 3 of the coins from the heavier side, move 3 coins from the lighter side to the heavier side,
......and add 3 of the coins that we know are the same to the lighter side, and measure again (measure #3)
......if the same, then it is one of the 3 coins that we took out, and we know the different one is heavier
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know it is heavier
.........if not the same, then it is the heavier one
......if the scale changed direction, then it is one of the 3 coins that we moved from the lighter side,
......and we know the different one is lighter
.........take 2 of them and compare them against one another (measure #4)
.........if the same then it is the coin that we didn't measure and we already know it is lighter
.........if not the same, then it is the lighter one
......if the scale remained the same, then it is one of the 2 coins (1 on each side) that we didn't move.
......Compare the heavier one against one coin that we know is the same as other (measure #4)
.........if the same then it is the one that we didn't measure and it is lighter
.........if not the same then it is the one that we just measure and it is heavier

I hope this was clear enough
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### #75 bhramarraj

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Posted 03 October 2011 - 09:07 AM

Spoiler for Not so harder way

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### #76 Ranbir Singh

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Posted 26 November 2011 - 04:17 PM

if you are not lucky , u will solve it in at max 6 tries...if u r somewhat lucky,u can do it in 5..and u can do it in 4 if u r very lucky...
IF U UNDERSTAND THIS IT MEANS YOU HAVE THE COMPLETE ANSWER.
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### #77 phillip1882

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Posted 26 November 2011 - 05:38 PM

Spoiler for the correct way to do it in 4

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### #78 ujjagrawal

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Posted 30 May 2012 - 12:20 PM

Atmost 4 measurements are enough, for 27 coins, equation goes like this
For m numbers of coin, where m>= 3^n (maximize n).
n measurements if it known whether different coin is heavier or lighter than other.
n +1 , if if you don't know if different coin is heavier or lighter than other.
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### #79 bhramarraj

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Posted 02 June 2012 - 09:18 AM

Spoiler for simple

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### #80 bhramarraj

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Posted 05 June 2012 - 10:00 AM

Sorry.... my solution at post no. 80 was not clear and with some errors. So as a penalty I am posting more clear solution... thanks
Spoiler for Whether lucky or unlucky

Edited by bhramarraj, 05 June 2012 - 10:05 AM.

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