rocdocmac

Certainly!

Very old one (2007/8). Not new!

@Thalia ... Done some counting, but still a lot to do! Nine combinations of 4-cubelet removal still to be done.

10 minutes each! 1. A bag holds four counters. One of them is white. Each of the others is either black or white at equal chance. One randomly draws out two counters, and discover they are both white. If one then randomly draws a third counter, what is the chance that it is white? 2. I’m about to play the music songs on Side 1 of a standard LP (diameter = 292 cm) that contains six tracks. The recorded surface area is 487.6 dm2 (including the “gaps” between each song) and the average distance between each annulus (i.e. distance between each “groove” along the LP radius) is 197 µm. The total time that it will take to play this side right from the beginning of the first song to exactly the end of the last (6th) song is 18:04 minutes. If the “needle” (stylus) is poised 12.5 mm directly above the beginning of the first song and 2 mm away from the edge of the record, how far does the “needle” travel to the point where the last song just ended? 3. What comes next? 6 1 3 1 4 _

Molly said: "Glass/Silicon (which is in the Carbon group!)"

Plotting the sequence 1, 4, 22, 139, ... as a scatter diagram, one can fit a 3rd degree polynomial curve with equation y = 14x^3 - 34.5x^2 + 23.5x + 1 (x = 0, 1, 2, 3). Using this graph as an estimator, one can guess that the number of shapes will be in the region of 439 if four are removed. Therefore the guestimate for 500 is not far out! T25 appears to be 9 (see attached image) and I agree that T26 is 1. The roots indicate the distance between the two cubelet-middle points in the grid (x = y = z = 1). Thus, would the sequence end with ..., 9, 1, 0 (the 0 indicating no shape at all)? Will determine T24 a bit later!

Thanks Thalia, Interesting twist in the tale, for sure! Yes, surely 26 cubelets removed leaves a single cubelet, but where was the remaining one positionally located/"fixed" before removal of the other 26? Should be one specific remnant shape for each combination and, therefore, four variants/labels/possibilties. "Nothing" or "no shape" may be regarded as a "blank" stacked cube. (1) What's your suggestion for the end of the sequence? (2) What do our other BrainDenners think? (3) How many distinct shapes can one expect for a 4-cubelet removal? Just a ball-park figure, perhaps!

@Thalia ...

@Izzy ... @Molly Mae ... you were halfway there!