IDENTIFY THE FAKE COIN

[BASKARAN R]

Suppose you have 30 coins identical in every way except one is either slightly lighter or slightly heavier than the other 29 coins. You can not find this difference by mere perception. You have a balance beam scale to weigh coins against each other. Using it only 4 times, can you figure out which is the counterfeit coin and whether it is lighter or heaver than all the other coins?

[Peter Weisz]

Divide the thirty coins into four groups of 9, 9, 9, and 3.

Label the 9 coin piles as 9a, 9b, and 9c.

Weighing number one is 9a vs. 9b.

If they balance, the odd coin is in either in 9c or in the pile of 3.

If so, then weighing Number 2 is 9a vs. 9c.

If they balance, then the odd coin is in the pile of 3.

Label the three coins the pile of 3 as 3a, 3b, and 3c.

Weighing number 3 is 3a vs. 3b.

If they balance, the odd coin is 3c.

In that case, weighing number four is 3c vs. any other coin to determine if the odd coin is lighter or heavier than the rest.

 

If after weighing No. 3, coin 3a does not balance against coin 3b, then one of them must be the odd coin.

Remove the lighter coin, let’s say it’s 3b, and place coin 3c on the scale against the remaining heavier one, 3a.

If they balance, then the odd coin is 3b and it is lighter.

If they do not balance, then the odd coin is 3a and it is heavier.

 

Now let’s go back and see what happens if the first weighing has a different outcome.

Weighing number one was 9a vs. 9b. Let’s say that they do NOT balance. That means that the odd coin is in either 9a or 9b.

Note which of the two is heavier. Let’s say 9a is heavier. 

Weighing number two is 9a vs. 9c. They will either balance or 9a will be heavier. There is no possibility that 9a will be lighter.

If 9a and 9c balance that means that the odd coin is in 9b and that it is lighter. 

If 9a is again heavier, that means that the odd coin is in 9a and it is heavier than the rest. Let’s say it’s in 9a and it’s heavier.

Now divide 9a into three groups of three coins each. 

Label them 3A, 3B, and 3C.

Weighing number three is 3A vs 3B.

If they balance then the odd (heavier) coin is in 3C.

If they do not balance, the odd (heavier) coin is in the heavier of 3A or 3B.

Take the group of the three that contains the odd coin and divide it into 3 individual coins: 1a, 1b, and 1 c.

The fourth weighing is 1a vs. 1 b.

If they balance the odd is is 1c and it is heavier.

If they do not balance, then the odd coin is the heavier of 1a and 1b.

 

Using this same methodology, the odd coin and its status may be determined no matter how each of the weighings turn out.

[telesphorelalancette]

If you put half of the coins on each side of the scale, the lighter side is the one with the fake coin. Then you remove half of the coins leaving you with 15 coins. You put one coin aside and weight seven coins on each side, if the scale is even, you have found the fake coin, if not, the lighter side is the one with the fake coin leaving you with 7 coins. You put one coin aside and weigh 3 coins on each side, again if the scale is even, you have found the fake coin, if not, you are left with 3 coins. Now do I need to tell what to do?

[BASKARAN R]

32 minutes ago, telesphorelalancette said:

If you put half of the coins on each side of the scale, the lighter side is the one with the fake coin.

The fake one could be heavier also!

[bonanova]

29 minutes ago, telesphorelalancette said:

Same procedure but reverse.

A procedure that permits different interpretations does not help. If the counterfeit coin is lighter, then it's on the lighter side. If it's heavier then it's on the heavier side. All you can conclude is that the counterfeit coin is on the balance scale. But since you're weighing all the coins, you already knew it was on the scale. So you've learned nothing.

[BASKARAN R]

On 4/2/2016 at 2:55 PM, telesphorelalancette said:

Same procedure but reverse.

I am sorry, you have misunderstood the problem. You cannot assume that the fake one is lighter or heavier and then proceed. It could be lighter or heavier which we don't know. We have to find out the fake one and also find out whether it is lighter or heavier.

[Wilson]

Perhaps...

 

starting point...3 bundles of 9, 3 coins aside?

[BASKARAN R]

9 hours ago, Wilson said:

Perhaps...

 

 

  Hide contents

starting point...3 bundles of 9, 3 coins aside?

I have solved a similar problem for 12 coins in 3 weights. That can be used as a hint and you can set aside 12 coins and proceed with 9 + 9!

 

[telesphorelalancette]

On 02/04/2016 at 5:24 AM, BASKARAN R said:

The fake one could be heavier also!

Still the same procedure. What's wrong with you? Lol

[Wilson]

 As before 9,9,9 & 3. Hope this makes sense.

Any two weighings of the 9's.....

If both balance then cfc is in 3 aside. 2 weighings of any of these 3 singles will solve.

If only one balances, then we will be able to see which 9 contains the cfc and also know if it's h or l.

...split this 9 into 3x3, since we know h or l,one balance reduces to the target group of 3.

....split this group into 3x1, as before h or l is known so one balance = bingo?

 

[BASKARAN R]

7 hours ago, Wilson said:

Any two weighings of the 9's.....

If both balance then cfc is in 3 aside. 2 weighings of any of these 3 singles will solve.

This part is fine!

 

 

7 hours ago, Wilson said:

 As before 9,9,9 & 3. Hope this makes sense.

 

  Hide contents

 

Any two weighings of the 9's.....

 

If only one balances, then we will be able to see which 9 contains the cfc and also know if it's h or l.

...split this 9 into 3x3, since we know h or l,one balance reduces to the target group of 3.

....split this group into 3x1, as before h or l is known so one balance = bingo?

 

 

If only one balences, the cfc is one among the 18 coins in the 2 pans. You cannot find the cfc in 2 remaining weighs!

 

16 hours ago, telesphorelalancette said:

Still the same procedure. What's wrong with you? Lol

[Wilson]

My logic may well be flawed. ??? ¥

If for the weighing of 2x9's...1 balaces (cfc not in these 18)...1 unbalanced (cfc is in the only 9 not in the balanced set , and as i know which side of the scale it is/was on, i know h or l too.

From 9 knowing h/l, can solve in two balances.

Where have i gone wrong?

This is a fun puzzle.

[TTZN]

divide 3x10 (let's say A,B,C).weigh A and B,

Process 1

if balance, fake coil is in C. Weighing C and (A or B) will find fake coil is lighter or heavier.

Divide C into 2x5 and weigh them and find abnormal group(D).

Divide D to 2x2 and 1, weigh 2 and 2  -> if balance ,1 is fake. If not, in this case you already understand the abnormal side,reduce each coil from each side of beam scale and you will find out the fake coil ,may be it is in your hand or still on beam scale,

Process 2

If no balance,weigh (A or B) and C will find fake coil group is lighter or heavier.Let's say fake coil is in A. Divide A (same as the bold sentence).

My Answer may be wrong.

Really sorry for my bad English and explanation.

[BASKARAN R]

On 4/6/2016 at 2:17 PM, TTZN said:

divide 3x10 (let's say A,B,C).weigh A and B,

Process 1

if balance, fake coil is in C. Weighing C and (A or B) will find fake coil is lighter or heavier.

Divide C into 2x5 and weigh them and find abnormal group(D).

Divide D to 2x2 and 1, weigh 2 and 2  -> if balance ,1 is fake. If not, in this case you already understand the abnormal side,reduce each coil from each side of beam scale and you will find out the fake coil ,may be it is in your hand or still on beam scale,

weigh 1: A and B balance; fake coin is in C (10 coins)

weigh 2: CONFIRM WHETHER FAKE COIN IS LIGHTER OR HEAVIER (comparing with A or B)

weigh 3: fake coin is in D (5 coins)

weigh 4: fake coin is one among 2 or 1 which is left out. 

If the fake one is one among the 2 we have to go for weigh 5!

On 4/8/2016 at 8:17 AM, Peter Weisz said:

Divide the thirty coins into four groups of 9, 9, 9, and 3.

Label the 9 coin piles as 9a, 9b, and 9c.

Weighing number one is 9a vs. 9b.

If they balance, the odd coin is in either in 9c or in the pile of 3.

If so, then weighing Number 2 is 9a vs. 9c.

If they balance, then the odd coin is in the pile of 3.

Label the three coins the pile of 3 as 3a, 3b, and 3c.

Weighing number 3 is 3a vs. 3b.

If they balance, the odd coin is 3c.

In that case, weighing number four is 3c vs. any other coin to determine if the odd coin is lighter or heavier than the rest.

 

If after weighing No. 3, coin 3a does not balance against coin 3b, then one of them must be the odd coin.

Remove the lighter coin, let’s say it’s 3b, and place coin 3c on the scale against the remaining heavier one, 3a.

If they balance, then the odd coin is 3b and it is lighter.

If they do not balance, then the odd coin is 3a and it is heavier.

 

Now let’s go back and see what happens if the first weighing has a different outcome.

Weighing number one was 9a vs. 9b. Let’s say that they do NOT balance. That means that the odd coin is in either 9a or 9b.

Note which of the two is heavier. Let’s say 9a is heavier. 

Weighing number two is 9a vs. 9c. They will either balance or 9a will be heavier. There is no possibility that 9a will be lighter.

If 9a and 9c balance that means that the odd coin is in 9b and that it is lighter. 

If 9a is again heavier, that means that the odd coin is in 9a and it is heavier than the rest. Let’s say it’s in 9a and it’s heavier.

Now divide 9a into three groups of three coins each. 

Label them 3A, 3B, and 3C.

Weighing number three is 3A vs 3B.

If they balance then the odd (heavier) coin is in 3C.

If they do not balance, the odd (heavier) coin is in the heavier of 3A or 3B.

Take the group of the three that contains the odd coin and divide it into 3 individual coins: 1a, 1b, and 1 c.

The fourth weighing is 1a vs. 1 b.

If they balance the odd is is 1c and it is heavier.

If they do not balance, then the odd coin is the heavier of 1a and 1b.

 

Using this same methodology, the odd coin and its status may be determined no matter how each of the weighings turn out.

 

Great! The arguments given go well. You have left discussing the case 9a and 9b balance, 9a and 9c do not balance!

[phaze]

Here goes

1st weigh = (12)-(12) 6

if it is unbalanced you can suspect 12 coins of being heavy (H) and 12 of being light (see 2nd weigh A)

if balanced you can eliminate 24 coins and instead weigh 6 unknown coins (see 2nd weigh B)

 

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2nd weigh A (4H 4L)-(4H 4L) 4H 4L

if it is unbalanced you can suspect 4 coins of being heavy on the heavy side and 4 of being light on the light side (see 3rd weigh A)

if it is balanced the remaining coins you suspect are 4 coins of being heavy and 4 of being light  (see 3rd weigh A)

2nd weigh B (2)-(2) 2

if it is unbalanced you suspect 2 coins of being heavy and 2 light (see 3rd weigh B)

if it is balanced you are left with 2 unknown coins (see 3rd weigh C)

 

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3rd weigh A (LLH)-(LLH) HH

if it is unbalanced you suspect 1 coins of being heavy and 2 light (see 4th weigh A)

if it is balanced you are left with 2 coins either of which is heavy (see 4th weigh B)

3rd weigh B (HL)-(HL)

Unless something has gone seriously wrong it will be unbalanced and you suspect 1 heavy coin and one light (see 4th weigh C)

3rd weigh C  (1)-(B) 1 compare with a coin you know is balanced (there should be 28 of them)

Unknown Coin goes down = it is heavy

Unknown Coin goes up = it is light

Balanced you need to check other coin (see 4th weigh D)

 

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4th weigh A  (L)-(L) H

if it is unbalanced the coin that went up is too light

if it is balanced the remaining coin is too heavy

4th weigh B (H)-(H)

the coin that went down is too heavy

4th weigh C (L)-(B) H compare suspect light coin with a coin you know is balanced

if light coin goes up it is too light

if light coin goes down the universe is about to implode and you should make appropriate arrangements

if balanced then the remaining coin is too heavy

4th weigh D (1)-(B) compare with a coin you know is balanced (there should be 29 of them)

Unknown Coin goes down = it is heavy

Unknown Coin goes up = it is light

 

[Buddyboy3000]

I think I got the right answer

Spoiler

1st weigh: (9U to 9U) with 12 coins leftover (12U)                                                                                    U = unknown coin

If equal, the fake coin is in the group of 12 leftover coins (go to A)                                                            N = normal coin

If uneven, we have 9H and 9L (go to B)                                                                                                     H = possibly heavier fake coin

2nd weigh: A: (9U to 9N) with 3 coins leftover (3U)                                                                                   L = possibly lighter fake coin

If equal, then the fake coin is the the 3 leftover (3U) (go to C)                                                                                                                                     

If the unknown side is heavier, it is in 9H (go to A)

If the unknown side is lighter, it is in 9L (go to B)

                   B: (9H to 9N) with 9L leftover

If the possibly heavier coins are heavier, the fake coin is in 9H (go to A)

If the possibly heavier coins are not heavier, it is in 9L (go to B)

3rd weigh: A: (3H to 3H) with 3H leftover

If one side if heavier, the fake coin is in that 3H (go to A)

If both sides are equal, then it is in the 3H leftover (go to A)

                   B: (3L to 3L) with 3L leftover

If one side is lighter, the fake coin is in that 3L (go to B)

If both sides are equal, it is in the 3L leftover (go to B)

                   C: (3U to 3N) with none leftover

If the unknown side is heavier, then the fake coin is in 3H (go to A)

If the unknown side is lighter, then it is in 3L (go to B)

4th weigh: A: (1H to 1H) with 1H leftover

If one side is heavier, that side is the fake coin

If both sides are equal, the leftover coin is the fake coin

                 B: (1L to 1L) with 1L leftover

If one side is lighter, that side is the fake coin

If both sides are equal, the leftover coin is the fake coin

 

Edited November 20, 2016 by Buddyboy3000
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[telesphorelalancette]

Same procedure but reverse.