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1. Gift wrapping the Puzzle blocks

So the problem is to fill a cubical box of side S with convex blocks having integral-length edges {L W H} whose values are taken without replacement from the set { 1, 2, 3, 4, 5, 6, ..., S-2, S-1, S } ? Or must they be taken from the set that excludes S? That is, can one of the blocks have an edge length equal to S?

Assuming you meant things of value (perhaps coins?) into the boxes, then ...
3. Gift wrapping the Puzzle blocks

The LxWxH block dimension seems to rule out Since OP does not it seems the smallest box would be But that requirement probably was intended. Also, I don't see that unique coloring imposes any limits beyond that of unique dimensions.
4. The JayCees

Jane, Janice, Jack, Jasper, and Jim are five high-school chums. Their last names are, in some order, Carter, Carver, Clark, Clayton, and Cramer. What are their full names? Here are some clues. Jasper's mom is deceased. In deference to an influential family member, the Claytons agreed that if they ever had a daughter they would name her Janice. Jane's parents have never met Jack's parents. The Cramer and Carter children have been teammates on several of the school's athletic teams. When he heard that Carver was going to out of town on the night of the school's Father and Son banquet, Cramer called Mrs. Carver and offered to "adopt" her son for the evening. But Jack's father had already asked him to go. The Clarks and Carters, staunch Republicans who are very good friends, were delighted when their children began dating each other.
5. Once more with the Maiden

@plasmid My bad. It's not polite to post a puzzle and then go dark for two months. Apologies. In the first version there was no acceleration cost for either contestant. They both could stop on a dime, turn, and resume at full speed instantly. So here I've added a cost for angular change of velocity (for the boat only) but none for linear acceleration. Before I finished my solution my hard drive fried. I replaced my computer but I lost my work. I'll share how far I got and maybe we can finish this off collaboratively.
6. Photochromism 326 BC

I sort of hesitate to relay a tongue in cheek report I read a while back that photochromic materials are not of recent discovery, but were actually known back in the time of Alexander the Great. A black substance could be ground into powder and dissolved in water. Alexander’s troops would soak strips of cloth torn from their togas in this solution and tie them around their wrists. As the sun rose, traversed the sky and then set, the treated cloth would change color, and by glancing at them his men could tell the approximate time of day. They called it Alexander’s Rag Time Band.
7. Swapped numbers

In each case is there a swap? Or are there longer cycles as well?
8. Diameters

Hi @CynPyn and welcome to the Den. Let's accept from this that Rmax is the unit diameter circle. Now imagine a rectangle with unit diameter (diagonal has length 1.) That can be made to fit into Rmax . The question is does every unit-diameter region into Rmax ?
9. Diameters

The diameter of a closed, topologically bounded region of the plane is the greatest distance between two points in the region. Example: the diameter of a rectangle is the length of its diagonal. Of all the regions whose diameter equals 1, one of them, call it Rmax, encloses the largest area. Can you prove, or disprove, that Rmax also encloses all other regions of diameter 1? That is, that all other regions of diameter 1 can be made to fit inside Rmax?
10. Once more with the Maiden

Previously, Maiden’s boat could change its heading instantaneously. Ogre’s heading could change only by virtue of following a circular path along the shore at his current speed. His rotational speed was thus far from infinite, and perhaps that disadvantage was unfair. So in this final puzzle iteration we’ll limit the boat’s linear speed to be f times that of Ogre, as before, but now we’ll also limit the boat’s angular speed to be never greater than g times Ogre’s top angular speed. A moment’s thought tells us that unless g is greater than unity the boat’s best strategy is to run at full speed from the center to the shore, keeping its initial bearing, no matter where on the shore Ogre initially stands. That is, never to turn the boat. That sucks for Maiden (e.g., she loses if Ogre initially stands at the boat's initial heading) and it sucks as a puzzle. So we’ll say the boat can change heading faster than Ogre can. For clarity we’ll set g = 2. We’ll implement that limit by giving the boat’s motor three discrete settings that can be switched instantly an unlimited number of times: clockwise (CW), full speed ahead (FSA), and counterclockwise (CCW.) In the two turning modes the boat turns but maintains its position; in FSA mode it moves forward but does not turn. Boat’s path is thus a succession of arbitrarily short line segments joined at angles of Maiden's choosing, with the time cost of the angle depending on its size. If the boat starts in the middle of the lake, how large must f now be for Maiden to escape? Edit: Extra credit (tough): If Ogre's top speed is 1 lake-radius per minute, and Maiden chooses the boat's initial heading at the center, what's her shortest time safely to shore?
11. Ten Islands

Yes it is possible. The Triborough Bridge in New York connects Manhattan, the Bronx and' Queens.
12. Kill Bill redux

Hmmm. I hadn't considered that. Is there a different solution if the murderer did not own the murder weapon?
13. The Maiden and the Ogre

Nicely done. As for adding some wrinkle to the problem, how about this? Suppose we have to add some time for her to get out of the boat before she starts to run. We could say the ogre must have an angular separation of s radians from the boat when it lands, and then minimize f. Hmmm. I'm guessing the same path just minimizes f to a slightly larger value. Hi @The Lonewolf Brand and welcome to the Den. That would mean the maiden's boat and the ogre have the same speed. She can escape from a faster ogre.
14. The Maiden and the Ogre

Interesting idea. How would we pose that question exactly? .
15. American Indian Artwork

What a deliciously challenging and intriguing puzzle.
16. The Maiden and the Ogre

If you think you've heard this one before, read it carefully. It's not the standard puzzle. A beautiful maiden sits in a boat at the center of circular lake. On shore waits an ogre anxious to have his way with her. Being an excellent sprinter she knows she can outrun and therefore escape the lumbering ogre if only she can land her boat safely. But should the ogre reach her landing point first, alas, all will be lost. The boat is propelled by a motor capable of only a fraction f of the ogre's speed. What is the minimum value of f that will permit the Maiden to escape?
17. What is my clock telling me to do?

That's some two-year-old. And yes, it's time to ..
18. Probability 101

@EventHorizon said, But once Plato says that "one die is a 4," it doesn't merely prune﻿﻿﻿﻿ out the ones without 4's. That’s the crux of it. EH is correct that this reads on Monty Hall. But more directly on the long-running Teanchy-Beanchy post (one of his two kids is a girl, what’s the probability he has two girls.) First people said has to be 1/2. Then others (including me) said (all that matters is that) it can’t be BB so it’s 1/3. Both wrong. so I made this one up to show that the informant’s algorithm has to be known. Nicely explained.
19. Probability 101

And so thought poor Aristotle until today's class. He knew the probability of two dice making seven, until his teacher told him the value of one of his dice. Then he reasoned it to be different. Ah, the magic of conditional probability, he thought. But then he reasoned further that it was not the knowledge of which value one of his dice had, for it did not matter whether that value was 1 2 3 4 5 or 6. It was seemingly only that it had a value. But what kind of conditional probability is that? Was he not already aware of that? One of his dice has a particular value. Six values engender six cases. In each particular case he reasons the probability of seven changes to a new value. Worse, there are no other cases. Therefore in every case it changes to a new value. How then could it have the value he originally imagined? So there's the question that lurks within the flavor text. Beneath the surface perhaps, but now fully revealed for all to ponder.
20. Ten Islands

@CaptainEd Brilliant.
21. Kill Bill redux

@flamebirde Yes. The gun’s owner did the deed. Implied but not stated. Good catch.
22. Trains that pass in the night

If it takes the 10-car Silver Streak passenger train twice as long* to pass the 93-car Big Boy steam freight train going in the same direction as it does going the opposite direction, what is the ratio of their speeds? *Meaning the time during which there is any overlap of the trains.

Yes.
24. Kill Bill redux

Now, what was it we were talking about? Oh yes ... wherrzzz Beeel? I'll tell you. You know why I tell you? Because Bill would want me to. But this time the old man was not talking to the impossibly gorgeous Beatrix Michelle Kiddo. It was to a guy. A guy who was on a mission to kill Bill. A guy with the name of Al, Jack, Joe or Tom. Which one? Well, it's your job to determine which. Here's what we know from police interviews that followed the discovery of Bill's cold body on a lonely stretch of country road late last night, along with the fatal gun that belonged to one of them ... the murderer. So read on and see where the evidence leads. But fair warning, exactly half of each of the suspects' four statements are lies. Al: I didn't do it. Tom did it. Sure I own a gun. Joe and I were playing poker last night when Bill was shot. Jack: I didn't do it. Al did it. Joe and I were at the movies last night when Bill was shot. Bill was shot with Joe's gun. Joe: I was asleep when Bill was shot. Al lied when he said that Tom killed Bill. Jack is the only one of us who owns a gun. Tom and Bill were pals. Tom: I've never fired a gun in my life. I don't know who did it. Joe doesn't own a gun. I never saw Bill until they showed me the body. Who was the rat who done poor Bill in?
25. Probability 101

Plato: Good morning, class, today's lesson is on probability. Aristotle: Fantastic. I'm headed to Vegas this weekend, and I can use some pointers. P: Curb your enthusiasm kid, this is serious stuff. Here, roll this pair of dice, but don't look at the the result. OK, good. Now without looking, tell me the probability that you rolled a seven. If you're going to play craps this is important. By the way, I can tell you that one of your dice is a four. A: Hmm... So I could have rolled 41 42 43 44 45 46 14 24 34 54 or 64, all with equal likelihood, with 34 and 43 making seven. That's a probability of 2/11. P: You're on a roll kid, now let's do it again. Great. Again without looking, what's the probability you rolled a seven? By the way, I can tell you one of your dice is a one. A: So I could have rolled 11 12 13 14 15 16 61 51 41 31 or 21, with 16 and 61 making seven. Hmm... it's the same as before - 2/11. P: And if I had told you one of your dice was a five? A: Well ... I guess it really doesn't matter what number you tell me. It will always come out the same. The probability will be 2/11. P: So what can we deduce from that? A: That the probability of two dice making seven is ... 2/11. But wait... Hey, you're not really Professor Plato, are you? P: No. I'm an insurance salesman. So ... what exactly is the probability of making seven?
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