bonanova

Kudos to TSLF. Nicely done!

Because you've run out of time when you reached the store.

Great job all.

What is the next (and final) number in this sequence? 6 2 5 5 4 5 6 3 7 _ ?

My answer is not very different from BobbyGo's, mainly to recognize two switches light one bulb from each of the two groups, thereby eliminating analysis of some non-occurring cases. It took me awhile to conclude the 1- and 3-switch bulbs were different, then the rest fell quickly in place. I'll share my BA with BG.

I think gavinksong's description was best. k-man, the guy should insert a tat saying "What about" above, and "don't you understand?" below. Looks like a proof of the Riemann Hypothesis.

My first language is English and I'm still learning it. My writing is OK but my speech needs work.

I think Nins_Leprechaun has it. Nice puzzle.

It's a paradox, more mathematical than physical, of course, so it lacks a "right" answer. I'm giving it to Phil for the "Black Hole" factor. I'm into cosmology at the moment.

You can't make a loop, without using one of the dots twice. . That is true but you can loop in your line as long you do not go over your line * * * * * * * * * * * * You can start at the red dot, make the loop, and then end at the blue dot. what you say is true but the rules allow you to loop within lines as long as you do not go over them Think of 100 circles of radii 1, 2, 3, ..., 100 centered at x = 1, 2, 3, ..., 100. They are mutually tangent at the origin. Any number - a countable infinity of them - could be mutually tangent. And none of them "go over" any of the others. Between circles 1 and 100 there are 98 circles that do not violate the OP. You would have to add a statement that once two curves are made to touch, a third such line is prohibited, in order to limit the curves that touch to only two. So, mentally I will add that statement, and completely agree.

If a domino is two squares joined at a common edge, then a hexomino is a similar plane figure comprising six squares. Since a cube has six square sides, it follows that certain hexominos can be folded into a cube. Here is one: Col 173 +----+----+----+ | | +----+====+----+ === =|= ==| ==== + ===+ === =| ===| ==== + ===+ ==== | ===| === =+----+ There are others, of course. So this puzzle has two(+) parts. Count them. Counting as a single hexomino all shapes that differ only by rotation and/or reflection, sketch all the hexominos that will fold into a cube. How many are there? Corral them. Now fit them, without overlap, inside a rectangle or square. What is the smallest perimeter of that enclosure? Extra credit: what is the smallest area of that enclosure?

I agree with plasmid (and Barcallica): (25/72)+(10/72)+3/16 = (15/56)+(10/56)+3/56 = 0.5

k-man, do you have an equation for that?

Modifed though, as the circle is within your circle. Also if it rotates from 12 to 3, 3 to 6, 6 to 9 and comes back to start, that is four rotations, right? Yes, modified; yet similar. Both require the consideration that traversing a circular path (inside or out) affects the coin/circle in the same way that a rotation does, as it traverses that path. No, not four rotations - see k-man's picture. A rotation occurs each time the point p returns to its initial azimuth. Not each time the point p touches the outer circle.

You can't make a loop, without using one of the dots twice.

More fun with infinity. Agree or disagree? A chain with finite number of links needs support for its topmost link, or the chain will fall. But the (much heavier) infinite chain will never fall because each link hangs securely on the one above it.

An oldie but goodie. Famous quote: FF? Who penned it?