#9 Weighings (Note: this answer relates to the version of the question at this link, where all fake coins are of uniform weight 1 unit away from the genuine coins.)

This answer assumes you can put zero coins on the right pan and get a readout of the actual weight of the coins on the left pan.

You can determine the weight of a fake coin in 2 weighings.

Let $f$ and $g$ be the weight of a fake coin and a genuine coin, respectively.

1. Put any coin on the left and nothing on the right. Record the weight $w$.
2. Put all 99 coins on the left and nothing on the right. Record the weight $W$.

You have 30 fake coins and 69 genuine coins. So:

• if $W = 99w \pm 30$ then the excess weight comes from the fakes, so $f=w \pm 1, g=w$; likewise,
• if $W = 99W \pm 69$ then $f=w, g=w \pm 1$.

Now, you want to find any fake coin. I take this to mean that you want to identify one of the 30 fake coins in the pile of 99 coins, and it doesn't matter which fake coin you identify. Assume $f>g$ (reverse the logic if $f<g$).

Keep the right pan empty.

Put a pile of 50 coins in the left pan. The weight will be $50g + x$, where $x$ is the number of fake coins in the pan. So the other pile has $30-x$ fake coins. Choose whichever pile has the greater number of fake coins, disregard the rest, and continue.

At each iteration, place half the remaining coins on the left pan and nothing on the right. Round up or down (it doesn't matter which) to get an integer number of coins.

In the worst case, the fake coins are always split evenly between your two piles. You will weigh 50 coins (including 15 fakes), then 25 (8 fakes), 13 (4 fakes), 7 (2 fakes), 4 (1 fake), 2 (1 fake), 1 (1 fake).

The last weighing is needed because you don't know which of the 2 from the previous weighing was the fake. At the last weighing, you start with 2 coins and weigh one of them. If the weight of the weighed coin is $f$, that's the fake. If it is $g$, the other coin is fake.

That's 7 weighings. Add the initial 2 (which determined $f$ and $g$) to get 9 weighings altogether.

9 Weighings

(Note: this answer relates to the version of the question at this link, where all fake coins are of uniform weight 1 unit away from the genuine coins.)

This answer assumes you can put zero coins on the right pan and get a readout of the actual weight of the coins on the left pan.

You can determine the weight of a fake coin in 2 weighings.

Let $f$ and $g$ be the weight of a fake coin and a genuine coin, respectively.

1. Put any coin on the left and nothing on the right. Record the weight $w$.
2. Put all 99 coins on the left and nothing on the right. Record the weight $W$.

You have 30 fake coins and 69 genuine coins. So:

• if $W = 99w \pm 30$ then the excess weight comes from the fakes, so $f=w \pm 1, g=w$; likewise,
• if $W = 99W \pm 69$ then $f=w, g=w \pm 1$.

Now, you want to find any fake coin. I take this to mean that you want to identify one of the 30 fake coins in the pile of 99 coins, and it doesn't matter which fake coin you identify. Assume $f>g$ (reverse the logic if $f<g$).

Keep the right pan empty.

Put a pile of 50 coins in the left pan. The weight will be $50g + x$, where $x$ is the number of fake coins in the pan. So the other pile has $30-x$ fake coins. Choose whichever pile has the greater number of fake coins, disregard the rest, and continue.

At each iteration, place half the remaining coins on the left pan and nothing on the right. Round up or down (it doesn't matter which) to get an integer number of coins.

In the worst case, the fake coins are always split evenly between your two piles. You will weigh 50 coins (including 15 fakes), then 25 (8 fakes), 13 (4 fakes), 7 (2 fakes), 4 (1 fake), 2 (1 fake), 1 (1 fake).

The last weighing is needed because you don't know which of the 2 from the previous weighing was the fake. At the last weighing, you start with 2 coins and weigh one of them. If the weight of the weighed coin is $f$, that's the fake. If it is $g$, the other coin is fake.

That's 7 weighings. Add the initial 2 (which determined $f$ and $g$) to get 9 weighings altogether.

#9 Weighings

This answer assumes you can put zero coins on the right pan and get a readout of the actual weight of the coins on the left pan.

You can determine the weight of a fake coin in 2 weighings.

Let $f$ and $g$ be the weight of a fake coin and a genuine coin, respectively.

1. Put any coin on the left and nothing on the right. Record the weight $w$.
2. Put all 99 coins on the left and nothing on the right. Record the weight $W$.

You have 30 fake coins and 69 genuine coins. So:

• if $W = 99w \pm 30$ then the excess weight comes from the fakes, so $f=w \pm 1, g=w$; likewise,
• if $W = 99W \pm 69$ then $f=w, g=w \pm 1$.

Now, you want to find any fake coin. I take this to mean that you want to identify one of the 30 fake coins in the pile of 99 coins, and it doesn't matter which fake coin you identify. Assume $f>g$ (reverse the logic if $f<g$).

Keep the right pan empty.

Put a pile of 50 coins in the left pan. The weight will be $50g + x$, where $x$ is the number of fake coins in the pan. So the other pile has $30-x$ fake coins. Choose whichever pile has the greater number of fake coins, disregard the rest, and continue.

At each iteration, place half the remaining coins on the left pan and nothing on the right. Round up or down (it doesn't matter which) to get an integer number of coins.

In the worst case, the fake coins are always split evenly between your two piles. You will weigh 50 coins (including 15 fakes), then 25 (8 fakes), 13 (4 fakes), 7 (2 fakes), 4 (1 fake), 2 (1 fake), 1 (1 fake).

The last weighing is needed because you don't know which of the 2 from the previous weighing was the fake. At the last weighing, you start with 2 coins and weigh one of them. If the weight of the weighed coin is $f$, that's the fake. If it is $g$, the other coin is fake.

That's 7 weighings. Add the initial 2 (which determined $f$ and $g$) to get 9 weighings altogether.

#9 Weighings (Note: this answer relates to the version of the question at this link, where all fake coins are of uniform weight 1 unit away from the genuine coins.)

This answer assumes you can put zero coins on the right pan and get a readout of the actual weight of the coins on the left pan.

You can determine the weight of a fake coin in 2 weighings.

Let $f$ and $g$ be the weight of a fake coin and a genuine coin, respectively.

1. Put any coin on the left and nothing on the right. Record the weight $w$.
2. Put all 99 coins on the left and nothing on the right. Record the weight $W$.

You have 30 fake coins and 69 genuine coins. So:

• if $W = 99w \pm 30$ then the excess weight comes from the fakes, so $f=w \pm 1, g=w$; likewise,
• if $W = 99W \pm 69$ then $f=w, g=w \pm 1$.

Now, you want to find any fake coin. I take this to mean that you want to identify one of the 30 fake coins in the pile of 99 coins, and it doesn't matter which fake coin you identify. Assume $f>g$ (reverse the logic if $f<g$).

Keep the right pan empty.

Put a pile of 50 coins in the left pan. The weight will be $50g + x$, where $x$ is the number of fake coins in the pan. So the other pile has $30-x$ fake coins. Choose whichever pile has the greater number of fake coins, disregard the rest, and continue.

At each iteration, place half the remaining coins on the left pan and nothing on the right. Round up or down (it doesn't matter which) to get an integer number of coins.

In the worst case, the fake coins are always split evenly between your two piles. You will weigh 50 coins (including 15 fakes), then 25 (8 fakes), 13 (4 fakes), 7 (2 fakes), 4 (1 fake), 2 (1 fake), 1 (1 fake).

The last weighing is needed because you don't know which of the 2 from the previous weighing was the fake. At the last weighing, you start with 2 coins and weigh one of them. If the weight of the weighed coin is $f$, that's the fake. If it is $g$, the other coin is fake.

That's 7 weighings. Add the initial 2 (which determined $f$ and $g$) to get 9 weighings altogether.

#$\not{9}$ 8 Weighings

This answer assumes you can put zero coins on the right pan and get a readout of the actual weight of the coins on the left pan.

You can determine the weight of a fake coin in 2 weighings.

Let $f$ and $g$ be the weight of a fake coin and a genuine coin, respectively.

1. Put any coin on the left and nothing on the right. Record the weight $w$.
2. Put all 99 coins on the left and nothing on the right. Record the weight $W$.

You have 30 fake coins and 69 genuine coins. So:

• if $W = 99w \pm 30$ then the excess weight comes from the fakes, so $f=w \pm 1, g=w$; likewise,
• if $W = 99W \pm 69$ then $f=w, g=w \pm 1$.

Now, you want to find any fake coin. I take this to mean that you want to identify one of the 30 fake coins in the pile of 99 coins, and it doesn't matter which fake coin you identify. Assume $f>g$ (reverse the logic if $f<g$).

Keep the right pan empty.

Put a pile of 50 coins in the left pan. The weight will be $50g + x$, where $x$ is the number of fake coins in the pan. So the other pile has $30-x$ fake coins. Choose whichever pile has the greater number of fake coins, disregard the rest, and continue.

At each iteration, place half the remaining coins on the left pan and nothing on the right. Round up or down (it doesn't matter which) to get an integer number of coins.

In the worst case, the fake coins are always split evenly between your two piles. You will weigh 50 coins (including 15 fakes), then 25 (8 fakes), 13 (4 fakes), 7 (2 fakes), 4 (1 fake), 2 (1 fake), 1 (1 fake).

The last weighing is needed because you don't know which of the 2 from the previous weighing was the fake. At the last weighing, you start with 2 coins and weigh one of them. If the weight of the weighed coin is $f$, that's the fake. If it is $g$, the other coin is fake.

That's 7 weighings. Add the initial 2 (which determined $f$ and $g$) to get 9 weighings altogether.

This can be optimised at the 4 coin stage by putting 1 in the left and 2 in the right. If the difference is $g$, the unweighed coin is fake. If the difference is $2g-f$, the coin in the left pan is fake. Otherwise the difference is $(g+f)-g$, with the fake coin being one of the two. Weigh one of the two to determine the fake coin. This brings the total down to 8 weighings.

#9 Weighings

This answer assumes you can put zero coins on the right pan and get a readout of the actual weight of the coins on the left pan.

You can determine the weight of a fake coin in 2 weighings.

Let $f$ and $g$ be the weight of a fake coin and a genuine coin, respectively.

1. Put any coin on the left and nothing on the right. Record the weight $w$.
2. Put all 99 coins on the left and nothing on the right. Record the weight $W$.

You have 30 fake coins and 69 genuine coins. So:

• if $W = 99w \pm 30$ then the excess weight comes from the fakes, so $f=w \pm 1, g=w$; likewise,
• if $W = 99W \pm 69$ then $f=w, g=w \pm 1$.

Now, you want to find any fake coin. I take this to mean that you want to identify one of the 30 fake coins in the pile of 99 coins, and it doesn't matter which fake coin you identify. Assume $f>g$ (reverse the logic if $f<g$).

Keep the right pan empty.

Put a pile of 50 coins in the left pan. The weight will be $50g + x$, where $x$ is the number of fake coins in the pan. So the other pile has $30-x$ fake coins. Choose whichever pile has the greater number of fake coins, disregard the rest, and continue.

At each iteration, place half the remaining coins on the left pan and nothing on the right. Round up or down (it doesn't matter which) to get an integer number of coins.

In the worst case, the fake coins are always split evenly between your two piles. You will weigh 50 coins (including 15 fakes), then 25 (8 fakes), 13 (4 fakes), 7 (2 fakes), 4 (1 fake), 2 (1 fake), 1 (1 fake).

The last weighing is needed because you don't know which of the 2 from the previous weighing was the fake. At the last weighing, you start with 2 coins and weigh one of them. If the weight of the weighed coin is $f$, that's the fake. If it is $g$, the other coin is fake.

That's 7 weighings. Add the initial 2 (which determined $f$ and $g$) to get 9 weighings altogether.