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Call the modular linear system below, "The Pi=E Problem".
31415*x_1+92653*x_2+58979*x_3+32384*x_4+62643*x_5=27182 (mod 14142)
38327*x_1+95028*x_2+84197*x_3+16939*x_4+93751*x_5=81828 (mod 13562)
5820*x_1+97494*x_2+45923*x_3+7816*x_4+40628*x_5=45904 (mod 37309)
62089*x_1+98628*x_2+3482*x_3+53421*x_4+17067*x_5=52353 (mod 50488)
98214*x_1+80865*x_2+13282*x_3+30664*x_4+70938*x_5=60287 (mod 1688)
44609*x_1+55058*x_2+22317*x_3+25359*x_4+40812*x_5=47135 (mod 72420)
"The Pi=E Problem" is to find a vector, vec x=(x_1,x_2,x_3,x_4,x_5)
in ZZ^5 that satisfies the equations above. The left hand side, right
hand side, and moduli integers appearing in "The Pi=E Problem" are
successive 5 digit blocks of the decimal representations of the irrational
numbers Pi, E, and Sqrt(2). The reader should beware that, for "The
Pi=E Problem", the most obvious "Search" space has size
Lcm(14142,13562,37309,50488,1688,72420)^5 approx 2.012243051*10^115
"The Pi=E Problem" is a constructive feasibility problem. The challenge
for the reader is to find a solution to "The Pi=E Problem" before I present
my solution on March 3, 2010. If you want a better idea of what I
am looking for, please visit 64 more example problems of a similar
nature, but with solutions here:
64 More Example Problems with Solutions
https://www.planetquantum.com/Solve/Level1/Den1/Page1.htm
These problems are easier than the hardest problems I can do, but
how hard are they for you? Are you up to this challenge? I will give
credit on this web site to the first person to email me a correct
solution before March 3, 2010.