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Topic: An Ingenious Cubic Puzzle (Read 963 times) |
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K Sengupta
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An Ingenious Cubic Puzzle
« on: Dec 17th, 2006, 6:51am » |
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Determine analytically the minimum value of a positive integer constant S such that the equation A*(B^3) - B^3 + A + B = S has precisely :
(i) Three distinct solutions in positive integers.
(ii) Four distinct solutions in positive integers.
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| « Last Edit: Dec 17th, 2006, 10:55am by K Sengupta » |
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ThudnBlunder
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Re: An Ingenious Cubic Puzzle
« Reply #1 on: Dec 17th, 2006, 11:22am » |
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Did anyone do your last such problem?
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ThudnBlunder
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Re: An Ingenious Cubic Puzzle
« Reply #3 on: Dec 18th, 2006, 8:51am » |
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Oh, I meant An All Possible Value Puzzle.
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_med ium;action=display;num=1162566342
I see that no one has. Even Eigenray diidn't finish it.  He must have got distracted by a harder one. 
There the variables were exponents. Therefore more intractable, IMHO.
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| « Last Edit: Dec 18th, 2006, 7:01pm by ThudnBlunder » |
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Barukh
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Re: An Ingenious Cubic Puzzle
« Reply #4 on: Dec 18th, 2006, 11:20am » |
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For the 3-solutions part, I get 200.
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Barukh
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Re: An Ingenious Cubic Puzzle
« Reply #5 on: Dec 19th, 2006, 4:46am » |
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...and for 4-solutions: 9300.
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Eigenray
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Re: An Ingenious Cubic Puzzle
« Reply #6 on: Dec 19th, 2006, 9:35am » |
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Barukh, I think you are off by one: 200 and 9300 have 4 and 5 solutions, respectively. I get record values at
S = 2, 4, 12, 200, 4008, 274404, 1663680,
going up to 7 solutions, but I don't know how to do it "analytically" (one could certainly check up through 200 by hand, but I don't know if that counts).
Basically, for each divisor (b+1) of S, check if a-1 = (S-(b+1))/(b3+1) is an integer. One only needs to check those b satisfying b3+1 <= S-(b+1), and then add the solution (a, b) = (1, S-1).
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Barukh
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Re: An Ingenious Cubic Puzzle
« Reply #7 on: Dec 19th, 2006, 10:39am » |
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on Dec 19th, 2006, 9:35am, Eigenray wrote:| Barukh, I think you are off by one: 200 and 9300 have 4 and 5 solutions |
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Yes, of course! I missed one simple solution for A=1!
Quote:| S = 2, 4, 12, 200, 4008, 274404, 1663680 |
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Interesting: my 4-solution (actually, 5-solution) number is not on your list! Must see how I miscalculated it.
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Eigenray
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Re: An Ingenious Cubic Puzzle
« Reply #8 on: Dec 19th, 2006, 12:16pm » |
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on Dec 19th, 2006, 10:39am, Barukh wrote:| Interesting: my 4-solution (actually, 5-solution) number is not on your list! Must see how I miscalculated it. |
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I guess you missed 4008, 5016, and 8040, as these all have 5 solutions. How did you get your answers?
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| « Last Edit: Dec 19th, 2006, 12:17pm by Eigenray » |
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Eigenray
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Re: An Ingenious Cubic Puzzle
« Reply #9 on: Dec 19th, 2006, 2:17pm » |
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Ah, I see. If there are n+1 solutions for S, then there are | hidden: | 1<b1<b2<...<bn < S-1 such that
S = bi+1 mod bi3+1.
Taking B = (bi) = (1,2,3,4) gives your answer 9300. This is certainly a good way to construct values of S with any given number of solutions, but verifying minimality is tricky. For example, S(1,2,3,11) = 4008 is not too surprising, since (2,3,4,12) have lots of factors in common, but S(1,2,3,10) = 5016 and S(1,2,3,9) = 8040 are also lower.
However, the problem only asked for 4 solutions, and this is much easier. We may easily solve S = 2 mod 2, 3 mod 9, and 4 mod 28, to get S(1,2,3) = 228. On the other hand, if S < 228, then S >= (b+1)+(b3+1) implies b < 7, so it's not too much work to find S(1,3,4) = 200 as the smallest S with 4 solutions. (In fact, if S < 200, then b<6, so we need only check (1,2,4), (1,2,5), (1,3,5), and (1,4,5). |
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Barukh
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Re: An Ingenious Cubic Puzzle
« Reply #10 on: Dec 19th, 2006, 11:24pm » |
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on Dec 19th, 2006, 2:17pm, Eigenray wrote:
That's exactly how I got my answers. I just missed the obvious solution bn = S-1.
Nice problem!
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