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If below limits/inhibits access to others or any such issue, let me know and I'll fix it up.

Hard:

Please note, I don't know how to do some notation. Hopefully my meaning is clear. I apologize in advance for incorrect notation.
1. Due to the taxes affecting the potential growth through each room, you must pass through all the doubling doors first. To do so requires a determinant path, and thus requires at least 1 question per room until all the doubling rooms are done. Note: As we are trying to find the smallest c, we must content with the smallest possible q.
2. As such, q will be equal to the number of y rooms, in this case 10.
3. The total penalty by the time you finish the just the y rooms will be:
yEn .102^n (summation)
Let's call this p.
4. With that, the max potential earnings by the time the y rooms are completed is:
12^n - p
5. We still need to factor in the losses incurred by completing all the x rooms, changing the formula to:
12^n - p - x.10
or t
6. Fred only needs to charge you enough to set you back to your starting amount, in this case 1, as a gain of 0 is not a gain. This looks like:
1 = t - c^y - x*.10
or
c = (ln(t - 1 - x*.10))/(ln y)
8. As y and x are specified, we can now solve for c, giving us the minimum c necessary:
c = (ln(818.4 - 1 - 1))/(ln 10) ~= 2.91190299605

P.S. Apologies for so many edits. Struggling with math today apparently.

Medium:

As far as I can tell, as there is no penalty for taking a liar's door, and because of the loss of an individual of the same type, you'll end up traversing through 10 liar's doors and 10 truth teller doors no matter what. And again, because of the lack of penalty, the order doesn't matter. Therefore, no questions are needed and my answer is:
0 questions

Easy:

Depending on the setup of the doors, I think it's either:
20! (Factorial) = 2,432,902,008,176,640,000
or
20? or 20E0 = 210

Very Easy:

As each initial path ends with 1 door, only those paths matter. Adding in the hatch leaves us with:
21 doors.

Extremely Easy:

I mean, he lives in the room that all paths lead to, thus the one room (other than the starting room) that every visitor will encounter. I think I'm just missing something here.

If below limits/inhibits access to others or any such issue, let me know and I'll fix it up.

Hard:

Please note, I don't know how to do some notation. Hopefully my meaning is clear. I apologize in advance for incorrect notation.
1. Due to the taxes affecting the potential growth through each room, you must pass through all the doubling doors first. To do so requires a determinant path, and thus requires at least 1 question per room until all the doubling rooms are done. Note: As we are trying to find the smallest c, we must content with the smallest possible q.
2. As such, q will be equal to the number of y rooms, in this case 10.
3. The total penalty by the time you finish the just the y rooms will be:
yEn .102^n (summation)
Let's call this p.
4. With that, the max potential earnings by the time the y rooms are completed is:
12^n - p
5. We still need to factor in the losses incurred by completing all the x rooms, changing the formula to:
12^n - p - x.10
or t
6. Fred only needs to charge you enough to set you back to your starting amount, in this case 1, as a gain of 0 is not a gain. This looks like:
1 = t - c^y - x*.10
or
c^y = t - x*.10 - 1
or
c = (t - 1 - x*.10)^(1/y)
8. As y and x are specified, we can now solve for c, giving us the minimum c necessary:
c = (818.4 - 1 - 1)^(.1) ~= 1.9551959979

P.S. Apologies for so many edits. Struggling with math today apparently.

Medium:

As far as I can tell, as there is no penalty for taking a liar's door, and because of the loss of an individual of the same type, you'll end up traversing through 10 liar's doors and 10 truth teller doors no matter what. And again, because of the lack of penalty, the order doesn't matter. Therefore, no questions are needed and my answer is:
0 questions

Easy:

Depending on the setup of the doors, I think it's either:
20! (Factorial) = 2,432,902,008,176,640,000
or
20? or 20E0 = 210

Very Easy:

As each initial path ends with 1 door, only those paths matter. Adding in the hatch leaves us with:
21 doors.

Extremely Easy:

I mean, he lives in the room that all paths lead to, thus the one room (other than the starting room) that every visitor will encounter. I think I'm just missing something here.

If below limits/inhibits access to others or any such issue, let me know and I'll fix it up.

Hard:

Please note, I don't know how to do some notation. Hopefully my meaning is clear. I apologize in advance for incorrect notation.
1. Due to the taxes affecting the potential growth through each room, you must pass through all the doubling doors first. To do so requires a determinant path, and thus requires at least 1 question per room until all the doubling rooms are done. Note: As we are trying to find the smallest c, we must content with the smallest possible q.
2. As such, q will be equal to the number of y rooms, in this case 10.
3. The total penalty by the time you finish the just the y rooms will be:
yEn .102^n (summation)
Let's call this p.
4. With that, the max potential earnings by the time the y rooms are completed is:
12^n - p
5. We still need to factor in the losses incurred by completing all the x rooms, changing the formula to:
12^n - p - x.10
or t
6. Fred only needs to charge you enough to set you back to your starting amount, in this case 1, as a gain of 0 is not a gain. This looks like:
1 = t - c^y - x*.10
or
c = (ln(t - 1 - x*.10))/(ln y)
8. As y and x are specified, we can now solve for c, giving us the minimum c necessary:
c = (ln(920.6 - 1 - 1))/(ln 10) ~= 2.96312644109

Medium:

As far as I can tell, as there is no penalty for taking a liar's door, and because of the loss of an individual of the same type, you'll end up traversing through 10 liar's doors and 10 truth teller doors no matter what. And again, because of the lack of penalty, the order doesn't matter. Therefore, no questions are needed and my answer is:
0 questions

Easy:

Depending on the setup of the doors, I think it's either:
20! (Factorial) = 2,432,902,008,176,640,000
or
20? or 20E0 = 210

Very Easy:

As each initial path ends with 1 door, only those paths matter. Adding in the hatch leaves us with:
21 doors.

Extremely Easy:

I mean, he lives in the room that all paths lead to, thus the one room (other than the starting room) that every visitor will encounter. I think I'm just missing something here.

If below limits/inhibits access to others or any such issue, let me know and I'll fix it up.

Hard:

Please note, I don't know how to do some notation. Hopefully my meaning is clear. I apologize in advance for incorrect notation.
1. Due to the taxes affecting the potential growth through each room, you must pass through all the doubling doors first. To do so requires a determinant path, and thus requires at least 1 question per room until all the doubling rooms are done. Note: As we are trying to find the smallest c, we must content with the smallest possible q.
2. As such, q will be equal to the number of y rooms, in this case 10.
3. The total penalty by the time you finish the just the y rooms will be:
yEn .102^n (summation)
Let's call this p.
4. With that, the max potential earnings by the time the y rooms are completed is:
12^n - p
5. We still need to factor in the losses incurred by completing all the x rooms, changing the formula to:
12^n - p - x.10
or t
6. Fred only needs to charge you enough to set you back to your starting amount, in this case 1, as a gain of 0 is not a gain. This looks like:
1 = t - c^y - x*.10
or
c = (ln(t - 1 - x*.10))/(ln y)
8. As y and x are specified, we can now solve for c, giving us the minimum c necessary:
c = (ln(818.4 - 1 - 1))/(ln 10) ~= 2.91190299605

P.S. Apologies for so many edits. Struggling with math today apparently.

Medium:

As far as I can tell, as there is no penalty for taking a liar's door, and because of the loss of an individual of the same type, you'll end up traversing through 10 liar's doors and 10 truth teller doors no matter what. And again, because of the lack of penalty, the order doesn't matter. Therefore, no questions are needed and my answer is:
0 questions

Easy:

Depending on the setup of the doors, I think it's either:
20! (Factorial) = 2,432,902,008,176,640,000
or
20? or 20E0 = 210

Very Easy:

As each initial path ends with 1 door, only those paths matter. Adding in the hatch leaves us with:
21 doors.

Extremely Easy:

I mean, he lives in the room that all paths lead to, thus the one room (other than the starting room) that every visitor will encounter. I think I'm just missing something here.

If below limits/inhibits access to others or any such issue, let me know and I'll fix it up.

Hard:

Please note, I don't know how to do some notation. Hopefully my meaning is clear. I apologize in advance for incorrect notation.
1. Due to the taxes affecting the potential growth through each room, you must pass through all the doubling doors first. To do so requires a determinant path, and thus requires at least 1 question per room until all the doubling rooms are done. Note: As we are trying to find the smallest c, we must content with the smallest possible q.
2. As such, q will be equal to the number of y rooms, in this case 10.
3. The total penalty by the time you finish the just the y rooms will be:
yEn .102^n (summation)
Let's call this p.
4. With that, the max potential earnings by the time the y rooms are completed is:
12^n - p
5. We still need to factor in the losses incurred by completing all the x rooms, changing the formula to:
12^n - p - x.10
or t
6. Fred only needs to charge you enough to set you back to your starting amount, in this case 1, as a gain of 0 is not a gain. This looks like:
1 = t - c^y - x*.10
or
c = (t + 1 - x*.10)^(1/y)
8. As y and x are specified, we can now solve for c, giving us the minimum c necessary:
c = (818.4 + 1 - 1)^(.10) ~= 1.9556744

Medium:

As far as I can tell, as there is no penalty for taking a liar's door, and because of the loss of an individual of the same type, you'll end up traversing through 10 liar's doors and 10 truth teller doors no matter what. And again, because of the lack of penalty, the order doesn't matter. Therefore, no questions are needed and my answer is:
0 questions

Easy:

Depending on the setup of the doors, I think it's either:
20! (Factorial) = 2,432,902,008,176,640,000
or
20? or 20E0 = 210

Very Easy:

As each initial path ends with 1 door, only those paths matter. Adding in the hatch leaves us with:
21 doors.

Extremely Easy:

I mean, he lives in the room that all paths lead to, thus the one room (other than the starting room) that every visitor will encounter. I think I'm just missing something here.

If below limits/inhibits access to others or any such issue, let me know and I'll fix it up.

Hard:

Please note, I don't know how to do some notation. Hopefully my meaning is clear. I apologize in advance for incorrect notation.
1. Due to the taxes affecting the potential growth through each room, you must pass through all the doubling doors first. To do so requires a determinant path, and thus requires at least 1 question per room until all the doubling rooms are done. Note: As we are trying to find the smallest c, we must content with the smallest possible q.
2. As such, q will be equal to the number of y rooms, in this case 10.
3. The total penalty by the time you finish the just the y rooms will be:
yEn .102^n (summation)
Let's call this p.
4. With that, the max potential earnings by the time the y rooms are completed is:
12^n - p
5. We still need to factor in the losses incurred by completing all the x rooms, changing the formula to:
12^n - p - x.10
or t
6. Fred only needs to charge you enough to set you back to your starting amount, in this case 1, as a gain of 0 is not a gain. This looks like:
1 = t - c^y - x*.10
or
c = (ln(t - 1 - x*.10))/(ln y)
8. As y and x are specified, we can now solve for c, giving us the minimum c necessary:
c = (ln(920.6 - 1 - 1))/(ln 10) ~= 2.96312644109

Medium:

As far as I can tell, as there is no penalty for taking a liar's door, and because of the loss of an individual of the same type, you'll end up traversing through 10 liar's doors and 10 truth teller doors no matter what. And again, because of the lack of penalty, the order doesn't matter. Therefore, no questions are needed and my answer is:
0 questions

Easy:

Depending on the setup of the doors, I think it's either:
20! (Factorial) = 2,432,902,008,176,640,000
or
20? or 20E0 = 210

Very Easy:

As each initial path ends with 1 door, only those paths matter. Adding in the hatch leaves us with:
21 doors.

Extremely Easy:

I mean, he lives in the room that all paths lead to, thus the one room (other than the starting room) that every visitor will encounter. I think I'm just missing something here.