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The answer is $8$ uses of the scale.

##First Weighing## Take one coin away and make $4$ groups of $30$. Pick any two of these groups and weight them. If scale balances, then they are all good coins. If the scale tips, then you know there is at least one bad coin in those $60$ and the other $61$ are good. Either way, you have eliminated at least $60$ coins in one weighing leaving you with either $60$ or $61$ coins.

##Second Weighing## WLOG, we will assume you have $61$ coins. Take one away and split into $4$ groups of $15$. Again, pick any two and weigh them. You will be able to eliminate at least $30$ more coins in this way.

Third Weighing##

You now have $30$ coins, plus one set aside, plus one in your pocket. The one set aside may or may not have been ruled out as a possible bad coin. Lets say that we still don't know, so that you have $31$ coins. Add your coin from your pocket to make $32$ and divide into $4$ groups of $8$. Weigh two groups, and you will be able to eliminate $16$, leaving you with at most $16$ coins.

##Forth Weighing## Split into $4$ groups of $4$. Eliminate $8$ by weighing two of the groups.

##Fifth Weighing## Split into $4$ groups of $2$. Eliminate $4$ by weighing two of the groups.

##Sixth Weighing## Weigh $2$ of the $4$ remaining coins. $2$ will be eliminated.

##Seventh and Eighth Weighings## Weigh the last $2$ coins. If the scale balances, you are done and all coins are equal. If the scale tips, then use your last attempt to see which one is the bad one by comparing one of them to a known good coin. You now have only $4$ coins left. Take $3$ of the coins and one from the pile of known good coins. Weigh 2 on each side.

The answer is $8$ uses of the scale.

First Weighing

Take one coin away and make $4$ groups of $30$. Pick any two of these groups and weight them. If scale balances, then they are all good coins. If the scale tips, then you know there is at least one bad coin in those $60$ and the other $61$ are good. Either way, you have eliminated at least $60$ coins in one weighing leaving you with either $60$ or $61$ coins.

Second Weighing

WLOG, we will assume you have $61$ coins. Take one away and split into $4$ groups of $15$. Again, pick any two and weigh them. You will be able to eliminate at least $30$ more coins in this way.

Third Weighing

You now have $30$ coins, plus one set aside, plus one in your pocket. The one set aside may or may not have been ruled out as a possible bad coin. Lets say that we still don't know, so that you have $31$ coins. Add your coin from your pocket to make $32$ and divide into $4$ groups of $8$. Weigh two groups, and you will be able to eliminate $16$, leaving you with at most $16$ coins.

Forth Weighing

Split into $4$ groups of $4$. Eliminate $8$ by weighing two of the groups.

Fifth Weighing

Split into $4$ groups of $2$. Eliminate $4$ by weighing two of the groups.

Sixth Weighing

Weigh $2$ of the $4$ remaining coins. $2$ will be eliminated.

Seventh and Eighth Weighings

Weigh the last $2$ coins. If the scale balances, you are done and all coins are equal. If the scale tips, then use your last attempt to see which one is the bad one by comparing one of them to a known good coin. You now have only $4$ coins left. Take $3$ of the coins and one from the pile of known good coins. Weigh 2 on each side.

Edit: Found an answer with 6 uses.

Take one coin away and split the rest into $3$ groups of $40$. Call them $A$, $B$, $C$. In your first two uses of the scale, you will weigh $A$ vs $B$ and $B$ vs $C$.

There are 7 possible results:

1. $A<B$ and $B=C$ - There is a light coin in $A$
2. $A>B$ and $B=C$ - There is a heavy coin in $A$
3. $A<B$ and $B>C$ - There is a heavy coin in $B$
4. $A>B$ and $B<C$ - There is a light coin in $B$
5. $A=B$ and $B<C$ - There is a heavy coin in $C$
6. $A=B$ and $B>C$ - There is a light coin in $C$
7. $A=B$ and $B=C$ - The coin set aside is the only unknown coin

After this is complete, you will know not only which $40$ coins has the faulty coin, but you will also know if it is lighter or heavier. If they are all the same, you can use the scale once more with the coin set aside to figure out if it is bad or good.

WLOG, lets assume you have $40$ with a single known heavy coin. Add $41$ known good coins to make $81$. Now, repeat the following:

• Split into three equal groups
• Weigh two of the groups
• You will know which of the three groups contains the coin

Until you are left with a single coin.

Since it took you 2 uses of the scale to get to $81$, here is how it breaks down:

3. $81$ coins becomes $27$
4. $27$ coins becomes $9$
5. $9$ coins becomes $3$
6. $3$ coins becomes $1$

So, you can do this in $6$ uses of the scale!

#Previous Answer#

The answer is $8$ uses of the scale.

The answer is $8$ uses of the scale.

The answer is $8$ uses of the scale.

Edit: Found an answer with 6 uses.

Take one coin away and split the rest into $3$ groups of $40$. Call them $A$, $B$, $C$. In your first two uses of the scale, you will weigh $A$ vs $B$ and $B$ vs $C$.

There are 7 possible results:

1. $A<B$ and $B=C$ - There is a light coin in $A$
2. $A>B$ and $B=C$ - There is a heavy coin in $A$
3. $A<B$ and $B>C$ - There is a heavy coin in $B$
4. $A>B$ and $B<C$ - There is a light coin in $B$
5. $A=B$ and $B<C$ - There is a heavy coin in $C$
6. $A=B$ and $B>C$ - There is a light coin in $C$
7. $A=B$ and $B=C$ - The coin set aside is the only unknown coin

After this is complete, you will know not only which $40$ coins has the faulty coin, but you will also know if it is lighter or heavier. If they are all the same, you can use the scale once more with the coin set aside to figure out if it is bad or good.

WLOG, lets assume you have $40$ with a single known heavy coin. Add $41$ known good coins to make $81$. Now, repeat the following:

• Split into three equal groups
• Weigh two of the groups
• You will know which of the three groups contains the coin

Until you are left with a single coin.

Since it took you 2 uses of the scale to get to $81$, here is how it breaks down:

3. $81$ coins becomes $27$
4. $27$ coins becomes $9$
5. $9$ coins becomes $3$
6. $3$ coins becomes $1$

So, you can do this in $6$ uses of the scale!

#Previous Answer#

The answer is $8$ uses of the scale.