bonanova

I find this easy to print out to work offline ... . A B C D E F G H I J K L M N O P Q R S T U V W X Y Z A . 3 1 2 1 1 2 3 1 0 1 2 2 2 3 2 3 3 2 1 3 1 2 1 2 0 B 3 . 0 2 3 2 3 3 3 3 2 2 2 1 2 3 2 3 0 3 1 2 1 2 3 2 C 1 0 . 1 1 1 1 1 2 2 2 2 3 4 2 1 1 2 4 1 4 2 2 2 2 1 D 2 2 1 . 1 1 2 1 2 1 1 2 2 1 4 4 2 1 2 2 2 2 0 1 3 2 E 1 3 1 1 . 3 4 2 2 4 2 2 3 3 1 1 2 3 3 2 2 1 2 4 2 3 F 1 2 1 1 3 . 4 3 1 4 2 4 3 3 1 1 2 2 3 1 3 0 2 4 2 2 G 2 3 1 2 4 4 . 3 1 3 1 3 3 3 2 2 2 3 2 3 2 2 2 3 3 3 H 3 3 1 1 2 3 3 . 2 2 1 4 4 3 2 2 2 2 2 3 2 1 3 3 2 1 I 1 3 2 2 2 1 1 2 . 2 1 2 2 1 3 2 1 2 1 3 1 3 0 3 2 2 J 0 3 2 1 4 4 3 2 2 . 3 2 2 2 1 1 1 2 2 2 2 1 1 4 2 3 K 1 2 2 1 2 2 1 1 1 3 . 1 1 2 0 1 2 3 1 1 3 1 2 3 3 2 L 2 2 2 2 2 4 3 4 2 2 1 . 5 4 2 3 2 2 3 2 3 2 3 3 3 1 M 2 2 3 2 3 3 3 4 2 2 1 5 . 5 2 3 2 2 3 2 3 2 4 3 3 1 N 2 1 4 1 3 3 3 3 1 2 2 4 5 . 1 2 1 3 5 1 4 2 4 3 3 1 P 3 2 2 4 1 1 2 2 3 1 0 2 2 1 . 2 2 0 2 2 2 2 0 1 2 2 P 2 3 1 4 1 1 2 2 2 1 1 3 3 2 2 . 2 2 1 2 2 2 2 1 4 1 Q 3 2 1 2 2 2 2 2 1 1 2 2 2 1 2 2 . 2 1 1 3 0 1 2 2 1 R 3 3 2 1 3 2 3 2 2 2 3 2 2 3 0 2 2 . 2 3 3 3 2 3 3 2 S 2 0 4 2 3 3 2 2 1 2 1 3 3 5 2 1 1 2 . 0 4 1 3 3 2 1 T 1 3 1 2 2 1 3 3 3 2 1 2 2 1 2 2 1 3 0 . 0 4 1 2 2 4 U 3 1 4 2 2 3 2 2 1 2 3 3 3 4 2 2 3 3 4 0 . 1 2 3 3 1 V 1 2 2 2 1 0 2 1 3 1 1 2 2 2 2 2 0 3 1 4 1 . 1 1 2 2 W 2 1 2 0 2 2 2 3 0 1 2 3 4 4 0 2 1 2 3 1 2 1 . 2 2 0 X 1 2 2 1 4 4 3 3 3 4 3 3 3 3 1 1 2 3 3 2 3 1 2 . 2 3 Y 2 3 2 3 2 2 3 2 2 2 3 3 3 3 2 4 2 3 2 2 3 2 2 2 . 1 Z 0 2 1 2 3 2 3 1 2 3 2 1 1 1 2 1 1 2 1 4 1 2 0 3 1 .

So let me ask: What is the minimum number of intersection lines for 5 points, no 4 of which are coplanar? [if the max and min are different.]

Toss out any three non collinear points and you've formed a triangle. Add a fourth non coplanar point. Each set of three points determines a triangle that intersects with the other triangles to form six lines which are the edges of a tetrahedron. Suppose now we add a fifth point such that no four of them are coplanar. Each set of three points determines a triangle that can extend to an infinite plane. What is the maximum number of lines that can result from the intersections of these planes?

That's information the prisoners don't need.

Be sure to count the first and last square. Or we could just say how many squares were not visited.

You got it ... my bad for not ruling out 144.

Did you read this, Professor? Y-san was looking at the lead story in today's Daily Blurb: "Vandals broke into the Flight of Fancy Bird Boutique last night, and more than half of their pet birds were stolen. Of the birds still in the store, 1/3 are parakeets, 1/4 are cockatiels, 1/7 are canaries, 1/8 are lovebirds and 1/9 are parrots." No, I didn't see that. Is that the whole story? No, there's more: "The Boutique had exactly 300 birds before the break-in, and the original number of canaries was double the number of parakeets still in their cages." The Professor took a puff on his pipe. Something doesn't seem right. Well, the reporter says he's not sure about the numbers. He thinks that one of his fractions might be wrong. Well, does it say how many canaries were stolen? No, Professor, but you hang around BrainDen a lot, why don't you ask someone there?

Imagine a 4x4 chess board with a lonely knight in the lower left corner. He decides to take a walk over to the lower right corner. A hop, actually; knights move in those silly L-shaped patterns. Being adventuresome, as knights are wont to be, he decides to visit as many squares as possible on his way, but without visiting any square twice. How many squares might he then visit, including his starting and ending squares?

Edit: Agree again. (1 + g) / c

I agree with the Professor

Lively debate about lots of topics, few if any of which qualify as a brain teaser. Won't be locked, but will be moved to Others.

Assuming the following We get these

Scratching my head. I think I've

Hi DE, and welcome to the Den. Since you didn't ask a question, the only response is a smile, from my school days. Because it's borderline non-PC, I'll leave it here to read but close off any more discussion. Looking forward to seeing some more riddles, and hoping you enjoy your time here. - bn

Hi UBUH, Sometimes an OP gives you some red herrings, and you get right to the answer by ignoring them. In this case, you ignored the expansion coefficient and the temperature change. That is a critical part of the puzzle, because it tells you how much longer the pipe became. You made the assumption that an arc is a semicircle and solved that case. But if it were, the pipe must have grown by a factor pi/2 - which does not agree with the information you ignored. Here's the place you went wrong ... any portion of a circle is an arc. Also, it's a good idea to put your solutions into a spoiler. That way members can choose whether to read it before solving it themselves.

Search "strongest +poison" Post answers here. Locked.

Hi Traffic, Doing the calculation is not that hard. If you understand what standard deviation is all about, the calculation even makes sense! So let's see what it is we're doing in the calculation. Start with a collection of numbers. You'd like to describe them somehow, perhaps to compare them with other collections of numbers. It could be the heights of your classmates, or how much money they have in their pockets. First thing you could do is find the average amount, called the mean. Add all the numbers, and get say $60.95. There are 23 classmates, say, so their average coin is $2.65. The mean is $2.65. What else might you want to know? Does every one have the same amount? Probably not. But if everyone doesn't have exactly $2.65, how much more or less do they have? Since some have more and some have less, the average of the differences is 0. Not helpful. But if you square the differences first, you get positive numbers, and the average won't be zero. So that's the idea. The average of the square of the differences from the mean gives you something called the variance. And its square root gives you the standard deviation. There's only one catch: When you take that last average, you don't use the number [23] of classmates, you use one less: 22. I won't go into why you do that, just believe: it gives a more meaningful result. So here's the deal: Standard Deviation: Find the mean. Find the differences from the mean. Square the differences. Find the average of the squares [but use N-1 instead of N] Take the square root. Now you have two ways to compare groups of numbers. The mean tells you the average amount of money owned by your classmates. The standard deviation tells you a kind of average of how much what they have differs from the mean. For example, if the standard deviation were zero, everyone would have exactly $2.65. If the standard deviation were huge, some of your classmates would have very little, and others quite a lot. Hope that helps. - bn

I spent a while on this and opted out for a rough approximation. This puzzle is a classic; sometimes named the Railroad Problem. The exact solution involves transcendental equations, easily solved now by computer, but close approximations can be made for an algebraic solution.

As promised ...

Uh ... no. But welcome to the Den - enjoy your time here.