I thought I had solved problems A & B, and had a very tentative solution re C which quite likely was wrong. C is clearly far more difficult and probably what I did on it should be regarded only as a start, not the whole kaboodle. But if other math funners work on it we may be able to solve C eventually. . I will here state the problems and explain my alleged solutions. Pizza hut does not seem to have announced solutions (?), and Conway's problems had some problems. OPTION A: I'm thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number? ANSWER: Assume wording was intended to be "for each n from 1 to 10, the number formed by the first n digits is divisible by n." Then the following are all 8-digit numbers that work, but only one can be extended to a 9-digit number by adding a digit to the right side, in such a way it still works: 12965408 30925864 36925840 38165472 72965480 78320416 78920456 80165472 92765408 08165432 so the resulting answer, which is unique, is: ** 3816547290 ** Also the answer obviously must end with 0, which makes it immediately clear all the above 8-digit guys are deadmeat except for the one. Unfortunately, Conway's problem A was NOT original, maybe he thought it was, but it wasn't. So therefore sombody who'd heard of it before, solved it in 1 minute by googling the answer. OPTION B: Our school's puzzle-club meets in one of the schoolrooms every Friday after school. Last Friday, one of the members said, "I've hidden a list of numbers in this envelope that add up to the number of this room." A girl said, "That's obviously not enough information to determine the number of the room. If you told us the number of numbers in the envelope and their product, would that be enough to work them all out?" He (after scribbling for some time): "No." She (after scribbling for some more time): "well, at least I've worked out their product." What is the number of the school room we meet in? ANSWER: cases where knowledge of number of numbers nn, sum su, and product pr fails to determine the full set of numbers (generated by computer): nn=3 pr=90 su=20 20=1+9+10 90=1*9*10 20=2+3+15 90=2*3*15 nn=3 pr=96 su=21 21=1+8+12 96=1*8*12 21=2+3+16 96=2*3*16 nn=3 pr=126 su=26 26=1+7+18 126=1*7*18 26=2+3+21 126=2*3*21 nn=3 pr=144 su=19 19=2+8+9 144=2*8*9 19=3+4+12 144=3*4*12 nn=3 pr=168 su=21 21=2+7+12 168=2*7*12 21=3+4+14 168=3*4*14 nn=3 pr=168 su=27 27=1+12+14 168=1*12*14 27=2+4+21 168=2*4*21 nn=3 pr=176 su=28 28=1+11+16 176=1*11*16 28=2+4+22 176=2*4*22 nn=3 pr=200 su=31 31=1+10+20 200=1*10*20 31=2+4+25 200=2*4*25 nn=3 pr=240 su=21 21=3+8+10 240=3*8*10 21=4+5+12 240=4*5*12 nn=3 pr=240 su=24 24=2+10+12 240=2*10*12 24=3+5+16 240=3*5*16 nn=3 pr=270 su=26 26=2+9+15 270=2*9*15 26=3+5+18 270=3*5*18 nn=3 pr=270 su=34 34=1+15+18 270=1*15*18 34=2+5+27 270=2*5*27 nn=3 pr=280 su=35 35=1+14+20 280=1*14*20 35=2+5+28 280=2*5*28 nn=3 pr=360 su=23 23=4+9+10 360=4*9*10 23=5+6+12 360=5*6*12 nn=3 pr=360 su=25 25=3+10+12 360=3*10*12 25=4+6+15 360=4*6*15 nn=3 pr=360 su=29 29=2+12+15 360=2*12*15 29=3+6+20 360=3*6*20 nn=3 pr=396 su=31 31=2+11+18 396=2*11*18 31=3+6+22 396=3*6*22 nn=3 pr=420 su=30 30=3+7+20 420=3*7*20 30=4+5+21 420=4*5*21 nn=3 pr=432 su=28 28=3+9+16 432=3*9*16 28=4+6+18 432=4*6*18 nn=3 pr=480 su=27 27=4+8+15 480=4*8*15 27=5+6+16 480=5*6*16 nn=3 pr=480 su=33 33=2+15+16 480=2*15*16 33=4+5+24 480=4*5*24 nn=3 pr=504 su=29 29=3+12+14 504=3*12*14 29=4+7+18 504=4*7*18 nn=3 pr=504 su=34 34=2+14+18 504=2*14*18 34=3+7+24 504=3*7*24 nn=3 pr=520 su=35 35=2+13+20 520=2*13*20 35=4+5+26 520=4*5*26 nn=3 pr=540 su=39 39=2+10+27 540=2*10*27 39=3+6+30 540=3*6*30 nn=3 pr=546 su=36 36=2+13+21 546=2*13*21 36=3+7+26 546=3*7*26 nn=3 pr=560 su=28 28=4+10+14 560=4*10*14 28=5+7+16 560=5*7*16 nn=3 pr=600 su=39 39=2+12+25 600=2*12*25 39=4+5+30 600=4*5*30 nn=3 pr=630 su=28 28=5+9+14 630=5*9*14 28=6+7+15 630=6*7*15 nn=3 pr=630 su=32 32=3+14+15 630=3*14*15 32=5+6+21 630=5*6*21 nn=3 pr=672 su=33 33=3+14+16 672=3*14*16 33=4+8+21 672=4*8*21 nn=3 pr=672 su=39 39=2+16+21 672=2*16*21 39=3+8+28 672=3*8*28 nn=3 pr=720 su=31 31=4+12+15 720=4*12*15 31=5+8+18 720=5*8*18 nn=3 pr=720 su=35 35=3+12+20 720=3*12*20 35=5+6+24 720=5*6*24 nn=3 pr=720 su=40 40=2+18+20 720=2*18*20 40=4+6+30 720=4*6*30 nn=3 pr=720 su=41 41=2+15+24 720=2*15*24 41=3+8+30 720=3*8*30 nn=3 pr=840 su=30 30=6+10+14 840=6*10*14 30=7+8+15 840=7*8*15 nn=3 pr=840 su=33 33=4+14+15 840=4*14*15 33=6+7+20 840=6*7*20 nn=3 pr=840 su=41 41=3+10+28 840=3*10*28 41=4+7+30 840=4*7*30 nn=3 pr=864 su=37 37=3+16+18 864=3*16*18 37=4+9+24 864=4*9*24 nn=3 pr=864 su=39 39=3+12+24 864=3*12*24 39=4+8+27 864=4*8*27 nn=3 pr=880 su=35 35=4+11+20 880=4*11*20 35=5+8+22 880=5*8*22 nn=3 pr=900 su=38 38=3+15+20 900=3*15*20 38=4+9+25 900=4*9*25 nn=3 pr=945 su=39 39=3+15+21 945=3*15*21 39=5+7+27 945=5*7*27 nn=4 pr=144 su=20 20=1+2+8+9 144=1*2*8*9 20=1+3+4+12 144=1*3*4*12 nn=4 pr=168 su=22 22=1+2+7+12 168=1*2*7*12 22=1+3+4+14 168=1*3*4*14 nn=4 pr=240 su=22 22=1+3+8+10 240=1*3*8*10 22=1+4+5+12 240=1*4*5*12 nn=4 pr=240 su=25 25=1+2+10+12 240=1*2*10*12 25=1+3+5+16 240=1*3*5*16 nn=4 pr=270 su=27 27=1+2+9+15 270=1*2*9*15 27=1+3+5+18 270=1*3*5*18 nn=4 pr=360 su=24 24=1+4+9+10 360=1*4*9*10 24=2+3+4+15 360=2*3*4*15 nn=4 pr=360 su=26 26=1+3+10+12 360=1*3*10*12 26=1+4+6+15 360=1*4*6*15 nn=4 pr=360 su=30 30=1+2+12+15 360=1*2*12*15 30=1+3+6+20 360=1*3*6*20 nn=4 pr=384 su=25 25=1+4+8+12 384=1*4*8*12 25=2+3+4+16 384=2*3*4*16 nn=4 pr=396 su=32 32=1+2+11+18 396=1*2*11*18 32=1+3+6+22 396=1*3*6*22 nn=4 pr=420 su=24 24=1+6+7+10 420=1*6*7*10 24=2+3+5+14 420=2*3*5*14 nn=4 pr=420 su=31 31=1+3+7+20 420=1*3*7*20 31=1+4+5+21 420=1*4*5*21 nn=4 pr=432 su=29 29=1+3+9+16 432=1*3*9*16 29=1+4+6+18 432=1*4*6*18 nn=4 pr=450 su=25 25=1+5+9+10 450=1*5*9*10 25=2+3+5+15 450=2*3*5*15 nn=4 pr=480 su=23 23=2+3+8+10 480=2*3*8*10 23=2+4+5+12 480=2*4*5*12 nn=4 pr=480 su=26 26=1+5+8+12 480=1*5*8*12 26=2+3+5+16 480=2*3*5*16 nn=4 pr=480 su=28 28=1+4+8+15 480=1*4*8*15 28=1+5+6+16 480=1*5*6*16 nn=4 pr=480 su=34 34=1+2+15+16 480=1*2*15*16 34=1+4+5+24 480=1*4*5*24 nn=4 pr=504 su=25 25=1+7+8+9 504=1*7*8*9 25=2+3+6+14 504=2*3*6*14 nn=4 pr=504 su=30 30=1+3+12+14 504=1*3*12*14 30=2+3+4+21 504=2*3*4*21 nn=4 pr=504 su=35 35=1+2+14+18 504=1*2*14*18 35=1+3+7+24 504=1*3*7*24 nn=4 pr=520 su=36 36=1+2+13+20 520=1*2*13*20 36=1+4+5+26 520=1*4*5*26 nn=4 pr=528 su=31 31=1+3+11+16 528=1*3*11*16 31=2+3+4+22 528=2*3*4*22 nn=4 pr=540 su=26 26=1+6+9+10 540=1*6*9*10 26=2+3+6+15 540=2*3*6*15 nn=4 pr=540 su=40 40=1+2+10+27 540=1*2*10*27 40=1+3+6+30 540=1*3*6*30 nn=4 pr=546 su=37 37=1+2+13+21 546=1*2*13*21 37=1+3+7+26 546=1*3*7*26 nn=4 pr=560 su=29 29=1+4+10+14 560=1*4*10*14 29=1+5+7+16 560=1*5*7*16 nn=4 pr=576 su=27 27=1+6+8+12 576=1*6*8*12 27=2+3+6+16 576=2*3*6*16 nn=4 pr=600 su=30 30=1+4+10+15 600=1*4*10*15 30=2+3+5+20 600=2*3*5*20 nn=4 pr=600 su=34 34=1+3+10+20 600=1*3*10*20 34=2+3+4+25 600=2*3*4*25 nn=4 pr=600 su=40 40=1+2+12+25 600=1*2*12*25 40=1+4+5+30 600=1*4*5*30 nn=4 pr=630 su=27 27=1+7+9+10 630=1*7*9*10 27=2+3+7+15 630=2*3*7*15 nn=4 pr=630 su=29 29=1+5+9+14 630=1*5*9*14 29=1+6+7+15 630=1*6*7*15 nn=4 pr=630 su=31 31=1+5+7+18 630=1*5*7*18 31=2+3+5+21 630=2*3*5*21 nn=4 pr=630 su=33 33=1+3+14+15 630=1*3*14*15 33=1+5+6+21 630=1*5*6*21 nn=4 pr=672 su=28 28=1+7+8+12 672=1*7*8*12 28=2+3+7+16 672=2*3*7*16 nn=4 pr=672 su=34 34=1+3+14+16 672=1*3*14*16 34=1+4+8+21 672=1*4*8*21 nn=4 pr=672 su=40 40=1+2+16+21 672=1*2*16*21 40=1+3+8+28 672=1*3*8*28 nn=4 pr=720 su=24 24=2+5+8+9 720=2*5*8*9 24=3+4+5+12 720=3*4*5*12 nn=4 pr=720 su=25 25=2+4+9+10 720=2*4*9*10 25=2+5+6+12 720=2*5*6*12 nn=4 pr=720 su=27 27=2+3+10+12 720=2*3*10*12 27=2+4+6+15 720=2*4*6*15 nn=4 pr=720 su=28 28=1+8+9+10 720=1*8*9*10 28=2+3+8+15 720=2*3*8*15 nn=4 pr=720 su=29 29=1+6+10+12 720=1*6*10*12 29=2+4+5+18 720=2*4*5*18 nn=4 pr=720 su=31 31=1+5+9+16 720=1*5*9*16 31=2+3+6+20 720=2*3*6*20 nn=4 pr=720 su=32 32=1+4+12+15 720=1*4*12*15 32=1+5+8+18 720=1*5*8*18 nn=4 pr=720 su=34 34=1+4+9+20 720=1*4*9*20 34=2+3+5+24 720=2*3*5*24 nn=4 pr=720 su=36 36=1+3+12+20 720=1*3*12*20 36=1+5+6+24 720=1*5*6*24 nn=4 pr=720 su=41 41=1+2+18+20 720=1*2*18*20 41=1+4+6+30 720=1*4*6*30 nn=4 pr=720 su=42 42=1+2+15+24 720=1*2*15*24 42=1+3+8+30 720=1*3*8*30 nn=4 pr=756 su=30 30=1+6+9+14 756=1*6*9*14 30=2+3+7+18 756=2*3*7*18 nn=4 pr=756 su=32 32=1+6+7+18 756=1*6*7*18 32=2+3+6+21 756=2*3*6*21 nn=4 pr=810 su=37 37=1+3+15+18 810=1*3*15*18 37=2+3+5+27 810=2*3*5*27 nn=4 pr=840 su=26 26=2+5+7+12 840=2*5*7*12 26=3+4+5+14 840=3*4*5*14 nn=4 pr=840 su=31 31=1+6+10+14 840=1*6*10*14 31=1+7+8+15 840=1*7*8*15 nn=4 pr=840 su=32 32=1+5+12+14 840=1*5*12*14 32=2+4+5+21 840=2*4*5*21 nn=4 pr=840 su=34 34=1+4+14+15 840=1*4*14*15 34=1+6+7+20 840=1*6*7*20 nn=4 pr=840 su=38 38=1+3+14+20 840=1*3*14*20 38=2+3+5+28 840=2*3*5*28 nn=4 pr=840 su=42 42=1+3+10+28 840=1*3*10*28 42=1+4+7+30 840=1*4*7*30 nn=4 pr=864 su=25 25=2+6+8+9 864=2*6*8*9 25=3+4+6+12 864=3*4*6*12 nn=4 pr=864 su=30 30=1+8+9+12 864=1*8*9*12 30=2+4+6+18 864=2*4*6*18 nn=4 pr=864 su=35 35=1+4+12+18 864=1*4*12*18 35=2+3+6+24 864=2*3*6*24 nn=4 pr=864 su=38 38=1+3+16+18 864=1*3*16*18 38=1+4+9+24 864=1*4*9*24 nn=4 pr=864 su=40 40=1+3+12+24 864=1*3*12*24 40=1+4+8+27 864=1*4*8*27 nn=4 pr=880 su=33 33=1+5+11+16 880=1*5*11*16 33=2+4+5+22 880=2*4*5*22 nn=4 pr=880 su=36 36=1+4+11+20 880=1*4*11*20 36=1+5+8+22 880=1*5*8*22 nn=4 pr=900 su=39 39=1+3+15+20 900=1*3*15*20 39=1+4+9+25 900=1*4*9*25 nn=4 pr=945 su=40 40=1+3+15+21 945=1*3*15*21 40=1+5+7+27 945=1*5*7*27 nn=4 pr=960 su=28 28=2+4+10+12 960=2*4*10*12 28=3+4+5+16 960=3*4*5*16 nn=4 pr=960 su=29 29=2+4+8+15 960=2*4*8*15 29=2+5+6+16 960=2*5*6*16 nn=4 pr=960 su=31 31=1+8+10+12 960=1*8*10*12 31=2+3+10+16 960=2*3*10*16 nn=4 pr=960 su=33 33=1+6+10+16 960=1*6*10*16 33=2+3+8+20 960=2*3*8*20 nn=4 pr=960 su=35 35=1+6+8+20 960=1*6*8*20 35=2+4+5+24 960=2*4*5*24 nn=4 pr=990 su=31 31=1+9+10+11 990=1*9*10*11 31=2+3+11+15 990=2*3*11*15 nn=5 pr=480 su=24 24=1+2+3+8+10 480=1*2*3*8*10 24=1+2+4+5+12 480=1*2*4*5*12 nn=5 pr=720 su=25 25=1+2+5+8+9 720=1*2*5*8*9 25=1+3+4+5+12 720=1*3*4*5*12 nn=5 pr=720 su=26 26=1+2+4+9+10 720=1*2*4*9*10 26=1+2+5+6+12 720=1*2*5*6*12 nn=5 pr=720 su=28 28=1+2+3+10+12 720=1*2*3*10*12 28=1+2+4+6+15 720=1*2*4*6*15 nn=5 pr=840 su=27 27=1+2+5+7+12 840=1*2*5*7*12 27=1+3+4+5+14 840=1*3*4*5*14 nn=5 pr=840 su=33 33=1+2+3+7+20 840=1*2*3*7*20 33=1+2+4+5+21 840=1*2*4*5*21 nn=5 pr=864 su=26 26=1+2+6+8+9 864=1*2*6*8*9 26=1+3+4+6+12 864=1*3*4*6*12 nn=5 pr=864 su=31 31=1+2+3+9+16 864=1*2*3*9*16 31=1+2+4+6+18 864=1*2*4*6*18 nn=5 pr=960 su=29 29=1+2+4+10+12 960=1*2*4*10*12 29=1+3+4+5+16 960=1*3*4*5*16 nn=5 pr=960 su=30 30=1+2+4+8+15 960=1*2*4*8*15 30=1+2+5+6+16 960=1*2*5*6*16 The only product in above list uniquely determined by sum is: sum=19, prod=144 from this line: nn=3 pr=144 su=19 19=2+8+9 144=2*8*9 19=3+4+12 144=3*4*12 It seems unreasonable for a school to have too many rooms in which case this search should have been exhaustive... anyhow it is certainly exhaustive if the numbers in envelope all are less than 32. The possibility of exceeding 32 does not hurt the validity of sum=19 as a solution. I have not fully proved 19 is the unique solution, but it quite likely is and others might want to try to show that. OPTION C: My key-rings are metal circles of diameter about two inches. (i) They are all linked together in a strange jumble, so that try as I might, I can't tell any pair from any other pair. (ii) However, I can tell some triple from other triples, even though I've never been able to distinguish left from right. What are the possible numbers of key-rings in this jumble? THOUGHTS AIMING TOWARD AN ANSWER: Let N=#rings. You said you had "rings" so N>1. You said you had "triples" so N>3. A pair of rings can be linked or not.

From (i) we see the linkedness graph G is either the complete graph K(N) or the no-edge graph (no two rings directly linked). (Clearly a pair which is linked, is distingushable from an unlinked pair.)

If all were unlinked, then there would have to be some subset whose removal would disconnect the whole jumble. If so there would exist some K pairwise-unlinked rings (perhaps with K<N), all linked, which would be disconnected by vanishing 1. That is impossible if K>2 due to M.Freedman & R.Skora: Strange Actions of Groups on Spheres, Journal of Differential Geometry 25 (1987) 75-98. Theorem: No Brunnian Link (for any N>2) can be built out of geometrically round circles. Here a "Brunnian link" is an N-loop link such that any two members are unlinked; and such that removing any single member disconnects it. They exist topologically for each N>2. The sole example when N=3 is Borromean rings. But it is not geometrically realizable. Therefore, it must be that case that either all N of our rings are pairwise linked, or: none are and removing N-3 causes remaining 3 still to be connected. The latter is impossible by Freeman-Skora and also [Bernt Lindstrom & Hans-Olov Zetterstrom: Borromean Circles are Impossible, American Mathematical Monthly 98,4 (Apr 1991) 340-341.] A triple of rings, all pairwise unlinked, can still be linked, but only as the "Borromean rings". But it is not possible for Borromean rings to be geometrical circles in 3-space, even when the radii are arbitrary. They could be ellipses, but problem said circles. OK, so only possibility is that all pairs of our circles are each pairwise linked. Our links, since N geometrical circles, must have crossing number=N*(N-1) if projected into any generic plane. And each component must cross each other exactly 2 times and never cross itself. Checking the Thistlethwaite Link Table, candidates are N=3, cross=6: exactly 3 candidates: L6a4=Borromean rings=unrealizable, L6a5=also unrealizable with circles in 3-space, seen by considering a sphere-inversion mapping the red circle to a line in http://katlas.org/wiki/File:L6a5.gif L6n1 works: http://katlas.org/wiki/File:Non-Borromean-rings_minimal-overlap2.png So if N>3 each triple must by itself be of topological type L6n1. Given two linked circles, the intersection of their planes, is a line. Given three linked circles, the intersection of their planes, is a point. It lies in the central region of the picture http://katlas.org/wiki/File:Non-Borromean-rings_minimal-overlap2.png A fourth circle in the jumble can be moved so its perimeter passes thru that point. Or not, if it went thru both the central South and Northernmost regions. That fact demonstrates that configurations with N=4 rings are of two distinguishable kinds, and anyhow not all triples are equivalent in the 2nd kind. But before proclaiming N=4 (second kind) as part of the answer we need to realize that the distinguishability of triples could imply some pairs are distinguishable. Since there are two kinds of triples in a foursome, that leads us to ask: find the incomplete N-vertex 3-regular hypergraphs H with the induced graph K(N) and such that all plain edges (2-vertex sets) are equivalent. H must be a "regular two-graph" ( https://en.wikipedia.org/wiki/Two-graph ). It must be one with doubly-transitive but not triply-transitive, automorphism group. A two-graph with N vertices exists iff there exists a nontrivial set of N "equiangular lines" in some dimensional space. Examples include N=6,10,16,28,36,126,176,344 according to http://oeis.org/A002853 . D.E Taylor: Two-graphs and doubly transitive groups, Journal of Combinatorial Theory, Series A, Volume 61, Issue 1, September 1992, Pages 113-122 Abstract Using the classification of finite doubly transitive groups the finite two-graphs which admit a doubly transitive group of automorphisms are determined. says the following N are the full list such that there exist two-graphs with doubly-transitive aut-groups for their N vertices (in the below, q denotes a prime power): N=q+1 where q=1 mod 4. (PSL acts 3-transitively if q=3 mod 4 but not if q=1 mod 4, says http://www.its.caltech.edu/~nbalacha/papers/paperfordcc.pdf ) N=q^3+1 with q odd. N=(8^n-2^n)/2 with n>2. N=(8^n+2^n)/2 with n>2. N=176. not triply trans. N=276. not triply trans. N=q^(2*n) where (n,q) is not (1,2). N=q^6. N=16 but the group A6 is 5-transitive. But that refers to 6D rep. N=64. Some of these N are redundantly generated such as 64, 16, q^6. So then a tentative answer to Conway's question is got by removing all the (>=3)-transitive cases from the Taylor list: N=q+1 where q=1 mod 4. N=q^3+1 with q odd. N=(8^n-2^n)/2 with n>2. N=(8^n+2^n)/2 with n>2. N=176. N=276. N=q^(2*n) where (n,q) is not (1,2). This is an infinite set of N. Now another flaw in Conway is physical. Due to the finite size of metal atoms, and the fact his rings are "about 2 inches" and universal linkitude forcing the diameter of the whole jumble to be < about 4 inches, only a finite number of these are physically possible; and even if we had magic metal with infinitesimal atoms, then still the Bekenstein entropy bound prevents more than a finite number of possibilities. He probably never thought of those issues. But also another problem in my whole argument above is there might be more than two kinds of triples (I was thinking only of 2 kinds, which we can view as "hypergraph arcs" and "not") -- and that needs to be understood. There was no way I was going to in only a few hours, which is all the time I had to work on this horror, so I left it there. But I think Conway's problems B and C were really masterpieces. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)