Suppose that M is an integer with the property that if x is randomly chosen from the set { 1, 2, 3, ..., 999,1000 }, the probability that x is a divisor of M...
Now that I know how to divide up a sphere into four equal pieces, using the tetrahedron to find the vertices, how can I general that to other numbers of equal...
Maybe it's a silly solution, but the watermellon idea "naturally" gives me this: with a regular dipyramid aka two pyramids glues by theis bases (bases are ...
Another way to state this method: Cut the watermelon the same way you would cut a circle into n congruent pieces. That is n radii with their outside ends...
In a message dated 6/5/2006 4:32:58 AM Central Standard Time, ... Drat! I forgot all about that. That's even the way I normally cut a watermelon. stevo ...
In a message dated 6/5/2006 5:04:00 AM Central Standard Time, ... 1005=a*100 + b*10 + c + a + b + c 1005= a*101 + b*11 + c*2 a=9 1005=909+b*11 + c*2 ...
... Yes, in lots of ways. :) I'm assuming pieces can overlap on their boundaries. If you disallow this, I'm pretty sure you get into open questions very...
... Or more succinctly: Find the largest integer M < 1000 that has exactly ten divisors (including 1 and M). If M = (p1^k1)*(p2^k2)*...(pn^kn) is the prime...
In a message dated 6/5/2006 11:44:49 AM Central Standard Time, ... So far I have f(990)=1008 f(CCC)=1019 base 13 f(666)=1023 base 7 f(AAA)=1027 base 11 I'm...
In a message dated 6/5/2006 12:23:59 PM Central Standard Time, ... Should be f(CCC)=1029 base 13. stevo [Non-text portions of this message have been removed]...
In a message dated 6/5/2006 12:23:59 PM Central Standard Time, ... f(888)=1025 base 9 f(AA9)=1025 base 11 ... In general the highest f can be in a given base b...
Stevo, We've sure got a lot of time on our hands.:-) Some decimal solutions n,m n = 100a+10b+c such that m = n+a+b+c Notice the missing values of m eg.,...
In a message dated 6/5/2006 2:56:02 PM Central Standard Time, ... Only by hand, so the answer is no. stevo [Non-text portions of this message have been...
... post ... face ... are no ... particular, ... congruent ... (snip) ... By the various methods that you describe I find several numbers of identical pieces...
... or ... Beautiful. That's a difficult exercise without actually making each ball. I think the one that dominates the page is an octahedron cut into sixes. ...
... You're perfectly correct; the rhombic dodecahedron has cubical (or octahedral) symmetry, but I missed it in my post because it arises by subdividing faces...
... but ... Good one. Also if the faces were not cut as mirror images, all the 3D pieces would also be identical; not half mirror images of the other half. ...
... but ... Good one. Also if the faces were not cut as mirror images, all the 3D pieces would also be identical; not half mirror images of the other half. ...
... When I saw this before I thought it was just a dodecahedron cut in fives; giving 60 pieces. Now I see that each triangle is joined to another forming 30...
... True, but then the result would be "polyhedral" only in two cases, when all the face-cutting rays pass through vertices, or through the edge midpoints. :) ...
... Perhaps surprisingly, same for the cube. A rhombic dodecahedron has 6 vertices incident on 4 faces, and 8 vertices incident on 3 faces. Bisecting rhombic...
If A, B and C are the angles of a triangle, r and R are respectively the inradius and circumradius of a triangle, then cos(A)+cos(B)+cos(C)=1+(r/R). This is an...
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