Close but not quite. 666*3 = 1998, so you have 2 left over.
If all three groups have exactly 333 heavy weights and 333 light weights, all three groups will weigh the same. (You meant put 2 aside, yes?)
This one had to be explained to me - I just couldn’t figure it out on my own:
What comes next in this series:
1, 11, 21, 1211, 111221, 11212211, 2112112221, …
[spoiler]The first number is arbitrarily 1. Each subsequent number describes the digits in the last. Understood best if you say each digit out loud. That is, the second number isn’t “eleven” it’s “one one” meaning the previous number consists of how many ones? one! So next is “two one” because “11” is two ones.
So the next number in the sequence is 122112213211[/spoiler]
The “gas” he bought is air, not gasoline. He can fix the problem by mending the puncture in his bicycle tire.
Er, ignore this post.
And what is the final number on the list? (odd or even)
Zero
The statement of problem 1 isn’t entirely clear to me, but:
[spoiler]It appears from the statement that you get the actual weight difference.
So put 666 vs 668. If the 668 is heaver by anything other than 198, 199, or 200 then just pull 2 off. If the 668 is heaver by 199, then switch one to the other side. If the 668 is heavier by 198 or 200, then the 666 is different than the 666 left out originally.[/spoiler]
A similar one. (Stolen from an episode of Columbo, where he went up against a Mensa member)
You have 10 sacks of gold ingots. One of the sacks contains fake ingots, the rest are real. The real ingots weigh 10 ounces, the fake ones weigh 11 ounces.
You have a set of accurate scales, which will tell you the weight of any objects placed in the pan. (Not a balance, just to be clear. No need to compare one ingot against another.)
How can you find the sack of fakes in the minimum number of weightings?
Assume that:
- the scales are accurate to 1/10th oif an ounce
- they are sturdy enough to hold any number of ingots
- you can’t tell the difference in weight simply holding them in your hands
- each sack contains as many ingots as you might need for the solution.
[spoiler] You can do it in one weighting:
Take 1 ingot from sack 1, 2 ingots from sack 2, 3 ingots from sack 3, etc.
If the weight is, for instance, 7 ounces over, then you know that sack 7 is fake [/spoiler]
Yes. Consider the regions along the circle inbetween fruit; there are 4n of these. Label each with (a - b)/2 where a is the number of apples and b is the number of bananas among the next 2n many fruit clockwise. Every region will be labelled with an integer, any two adjacent regions will be labelled by either the same or adjacent integers, and opposite regions (in the sense that each is the 2n-th region after the other) will have the negation of each other’s values; thus, there must be a pair of opposite regions labelled with zeros, and drawing a line through one point from each will solve the problem.
Oops!
Car Talk had a similar one some years back, except this time, you’re working with rolls of coins (100 coins per roll), and you have 7 rolls, and the difference between real and fake is 1 gram vs. 1.1 grams, and instead of knowing that one sack of ingots is fake, any or all of them could be fake. What’s the minimum number of weighings you need to to do figure out which, if any, rolls are fake? And then they told you: one weighing. The question is, how?
Ah, yes. Got it.
[spoiler]Use binary.
Take from each roll the following
r1 - 1
r2 - 2
r3 - 4
r4 - 8
r5 - 16
r6 - 32
r7 - 64
Then if the weighing is, for instance, 4.3 grams over, then that can only happen if r6, r4, r2, r1 are fake, and no others.
[/spoiler]
No, I simply was sloppy.
Here’s how it should work:
-
Divide the blocks into three groups, containing 666, 666 and 668 - call these G1, G2 and G3.
-
Weigh G1 and G2; if unequal you are done.
-
If equal, set G1 aside and move one block from G3 to G2 (which now contain 667 blocks each).
-
These two groups cannot have the same weight. Here’s why:
At the start of step #3, G1 and G2 can be of equal weight only if they had the same distribution of heavy (H) and light (L) weights. So G3.H (number of heavy weights in G3) must initially be equal to 1000 - 2 * G1.H. In other words, G3.H must be an even number.
If G1.H is odd, then G2.H plus G3.H must also be odd, so no matter what block is moved, there is no way that G2 and G3 can have an equal distribution of L and H.
How about if G1.H is even? Well, if G1.H = 332, then G3.H starts out as 336, so G2.H and G3.H can’t finish any closer than 333/335.
Q.E.D.
Shouldn’t the 6th number be 312211 if threes are allowed? Seems strange that it would be 1 one + 2 ones instead of 3 ones? If threes aren’t allowed, then I guess the next number would be 12211221122211, or alternatively, 12211221221211 
You are correct, especially since he does go up to 3’s in his 8th number.
One of my favourite “men in hats” puzzles:
Three men are in separate rooms. Someone places a white or back hat on each of their heads. They cannot see their own hats. They can observe each other on video monitors, but not communicate in any way.
They are given the option of guessing the colour of their own hat, or saying nothing. They do this simultaneously.
If anybody guesses incorrectly, or nobody guesses at all, they are naturally executed, that being the custom in Puzzleland. Otherwise, they are set free, or marry the king’s daughters, you know the drill.
I.e. they need at least one person to guess correctly and nobody to guess incorrectly.
Before this ordeal, they are allowed to agree a strategy. How can they maximise their chances of survival?
I think I must be missing a restriction here.
Can they identify who each other are on the video monitors?
If so, then it’s pretty easy:
[spoiler]Alan tells Bob and Carl to not guess, and tells Bob to raise his left hand if Alan is wearing a white hat, or his right hand if Alan is wearing a black hat. Then Alan guesses based on what Bob does on the video monitor.
If the video monitors only show the heads (but they are identifiable) then it could be shaking/nodding the head as the sign.[/spoiler]
No communication allowed. Maybe I should have said that they are in isolated cells and simply told what colour hats the other two are wearing.
Is anything known about the color distribution of the hats? Because if there’s nothing known, the best I can come up with is for one person to be the designated random guesser. If the hats are known to be mixed-color then the answer would be:
don’t guess if you are told two different colors of hats; if you are told two same-color hats, guess the opposite for yours.
Come to think, my spoilered strategy above would be better than designated-random-guesser even if the hats are randomly chosen and 3-black or 3-white is possible.