# Logic Puzzles

Ike, of course the man has a head. But he’s wearing someone else’s head that day.

Actually, he’s afflicted with the same syndrome Mike Nichols has, and he has no hair anywhere on him. This should take care of niggling details like eyelashes, eyebrows, etc.

Jabba, I’ve got you on #3. But only if you start by assuming that only the balsam is to be divided, 'cause I can’t even begin to evenly divide the containers equally.

Fill the 13 ounce and the 5 ounce from the original. This leaves 6 ounces in it.
Fill the 11 ounce from the 13 ounce. This leaves 2 ounces in the 13 ounce. Pour it into the original, and it’s 8 ounces.

Pour the 5 ounce into the 13 ounce. Fill the 13 ounce to the top from the 11 ounce. This leaves you with 13 ounces in the 13 ounce and 3 ounces in the 11 ounce.
Fill the 5 ounce from the 13 ounce. This leaves 8 ounces in the 13 ounce.

Pour your 5 ounce into the 11 ounce.

Your original container has 8 ounces.
Your 13 ounce container has 8 ounces.
Your 11 ounce container has 8 ounces.
You can do as you please with the 5 ounce container.

Yeah, he IS bald, and pigeon-toed as well, but that doesn’t explain why he only punches the 6th floor button.

And I am extremely tardy in answering.

He’s bald?

Swimming With Chickens suggested ( my bolding):

No good. The husband who goes back has left his wife with another man.

For Max Torque’s puzzle, the key is in the first weighing.

Ike: Because he’s a midget, and the 6th floor button is the highest one he can reach. Actually, I’ve heard that one with a bit more information: “Every day, he rides the elevator down to the lobby and goes to work. Every evening, he takes the stairs back up to his apartment, unless it’s raining, or if his wife is with him.” See, if it’s raining, he can reach the button with his umbrella…

I’m sticking with a headless horseman (on his way to the stable), who can’t reach the 12th floor button, but can reach the 1st floor button.

Or a midget, but only if we can say that there is a headless horseman midget.

Or someone without forearms. Can’t reach the 12 floor button without forearms.

I’ll call the husbands A, B, and C, and the wives A’, B’, and C’.

First trip, A’ and B’. B’ brings the boat back. Bank 1 is now A, B, C, B’, and C’; bank 2 is A’.
Second trip, B’ and C’. C’ brings the boat back. Bank 1 is now A, B, C, and C’; bank 2 is A’ and B’.
Third trip is A and B. Here is the key step: Both A and A’ bring the boat back. Bank 1 is now A, C, A’, and C’. Bank 2 is B and B’.
Fourth trip is A and C. At this point, we have A, B, C, and B’ on bank 2 with the boat. B’ can now ferry the other two across.

Actually, I think the key is in the second weighing.

Yes, I suppose in the case where the first weighing is the least informative it can be, you do need another cunning idea on the second weighing.

He’s an unmarried midget with no forearms who can’t afford an umbrella!

is that the real answer? cause if it is that sucks…

No, *Max Torque had the right answer. I was merely engaging in lighthearted word-play.

You weak-minded fool!!!

This puzzle is actually perfectly solvable if you have not twelve, but THIRTEEN coins. However, it’s easy to prove mathematically that thirteen is the maximum. (Interestingly, proving that 13 is the maximum is much easier than actually solving the problem for 13.)

As stated, this problem is easily demonstrable to be unsolvable, because there’s no way to possible distinguish the case where all 20 are aluminum from the case where 0 are aluminum.

But that’s a pretty cheap objection… I’ll have to think about it.

This handy little website goes through the steps of the logic puzzle quite well. Damn it’s a complicated one tho!

Well, that solution appears solid, but here’s my method. We’ll say the coins are numbered 1-12.

First Weighing: 1 2 3 4 against 5 6 7 8. Note which side is heavier and which is lighter.

Second Weighing: 1 2 3 5 against 4 9 10 11. On this weighing, there are three possibilities: the scale tips the same way, the scale balances, or the scales tip in the opposite way.

If the scales tip the same way, that is, if the side with “1 2 3” remains heavy or light on this weighing, you know that the counterfeit is either 1, 2, or 3, and you’ll know whether it is heavy or light. For the final weighing, weigh 1 against 2. If the scales tip, you’ll know which is the counterfeit, because you already figured out if the counterfeit is heavy or light; if they don’t tip, you know the counterfeit is 3.

If the scales balance, you’ll know that the counterfeit is one of the coins you removed (6, 7, or 8), and you’ll know whether it is heavy or light from the first weighing. For the final weighing, weigh 6 against 7. If the scales tip, you’ll know which is the counterfeit, because you already figured out if the counterfeit is heavy or light; if they don’t tip, you know the counterfeit is 8.

If the scales tip in the opposite way, then the counterfeit must be either 4 or 5. Changing the positions of these two coins caused the scale to tip in the opposite way. For the final weighing, weigh 4 against 1, because you know that 1 is not the counterfeit. If the scales unbalance, 4 is the counterfeit; if they don’t, 5 is the counterfeit (you won’t have figured out if it’s light or heavy, but you didn’t have to in this scenario, and your task was to find the counterfeit, not to figure out if it’s lighter or heavier).

But what if the scales balance on the first weighing, but unbalance on the second? Then you know that the counterfeit is one of the coins you added (either 9, 10, or 11), and you’ll know whether it is heavy or light. For the final weighing, weigh 9 against 10. If the scales tip, you’ll know which is the counterfeit, because you already figured out if the counterfeit is heavy or light; if they don’t tip, you know the counterfeit is 11.

And there’s a final possibility: if the scales balance for both the first and second weighings, coin 12 is the counterfeit, and a third weighing isn’t even necessary. But for giggles, you can weigh it against one of the good coins and find out if it’s heavy or light.

The trick, I feel, is in the second step: taking a coin from one side and swapping it with a coin on the other. That fixes it so that by the final step, no matter what, you have it narrowed down to at most 3 coins.

A young man walks into a professor’s office and asks for his daughter’s hand in marriage. The professor says, “I thought this day would come. You may marry my daughter if you prove your intelligence. Observe on the wall behind you are four light switches labeled one through four. These switches work four lamps on a table in my laboratory across the hall. You cannot see in the lab from the office or vice versa. The light switches are at the moment all turned off. You may stay here as long as you like and turn the switches on and off as often as you like. However, once you leave the office you may not return. The lab is directly across the hall and can be entered in seconds. Once you enter the laboratory, you must tell me which lamps are worked by which switches.”

How does our hero accomplish the task and marry the girl?

Let’s call the light switches A, B, C, and D.

Turn on A and B. Wait half an hour. Turn off B and turn on C. Go to lab immediately.

One lamp is on and the bulb is warm = A
One lamp is off but the bulb is still warm = B
One lamp is on but cold, because it’s only just been turned on = C
One lamp is off and cold = D

Oh, I guess I should post a puzzle … I wrote this one for a lateral thinking puzzle forum that allowed unlimited yes-or-no questions, so feel free to ask away.

Steve goes to the grocery store. He has a shopping list his wife has given him for a dinner party they are planning, which is complete and correct in every way. He buys every item on the list and leaves the store. Five minutes later he returns to the store and buys a large quantity of carrots, which were not on the list. He and his wife do not like carrots, nor do they intend to serve them to their guests or anyone else.

What are the carrots for? (No, they are not planning to use them as a sex toy; I guess I should head that one off before anyone asks.)