Problems & Puzzles: Puzzles

Puzzle 475. A question related to ABC Conjecture Jaroslaw Wroblewski sent the following puzzles: Find positive integers A, B, C with the following properties: A<B<C<10^1000 A+B=C GCD(A,B,C)=1 R = product of the distinct prime factors in (A*B*C) Find A, B, C such that the ratio C/R is as large as possible * Note that there is no log in the above ratio, but instead there is a limitation to C<10^1000. JW has found a solution such that C/R=3.488 * 10^8. Q. Can you improve the JW solution?

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J. K. Andersen wrote (Jan 09):

http://www.math.leidenuniv.nl/~desmit/abc/  shows a triple by
Ismael Jiménez Calvo using data from ABC@home :
A=23^8*37^4, B=2^28*3^7*11^4*19^3*61*127*173^2, C=5^18*17^4*43^2*4817^2.
C has 29 digits. C/R = 5.428*10^8.

See this page & this table related to Mr. Jiménez Calvo.

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In the meanwhile J. Wroblewski sent the following notice:

Here is the best example I was able to produce:

B = 3^226 * 11^180 * 29^49 * 37^24 * 41^254 * 43^41 * 47^10 * 53^45
C = 2^878 * 5^163 * 7^204 * 13^23 * 17^83 * 19^74 * 23^14 * 31^66 * 59^48

Assuming squarefree A=C-B (the only fair assumption when I cannot factor it) we have C/R = 7.79893958 * 10^22

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Question: From the JW last example, is really A square free?

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Anton Vrba wrote (Feb 09)

A =23⁸37⁴ B= 2²⁸3⁷11⁴19³61·127·173² C= 5¹⁸17⁴43²4817²
this 29 digit record tripple was found by Ismael Jiménez Calvo which yields a C/R = 5.428·10⁸


Amazingly, the above tripple was calculated from a lesser tripple
a = 23⁴37² b = 5⁹17²43.4811 c = 2²⁷19³·127
and multiplying each term by (a-b) resulting in the record tripple a2 + c(a-b) = b2

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