Theorem 2 Under the notation of Theorem 1, let gl=M0M0−1(−1)hh(−1)5hl(0)+M1M1−1g1h1(1)+⋯+MrMr−1grhr(r),−1⩽l⩽r,
where we take hl(D)=1,hl(j)=0,j=l,−1⩽j⩽r
Then g~−1≡−1(mod2α0),g0≡5(mod2a0)gj≡gj(modpjαj),1⩽j⩽r,
and the following φ(m) numbers form a complete set of reduced residues modulo m: {x=g−1r−1(1)g0(0)g1(1)⋯gr(r)0⩽γ(j)<cj,−1⩽j<r
Solution
None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". Here is the formatted output as requested:
None
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.
Source: NuminaMath-1.5,
licensed Apache-2.0.
Statement and solution reproduced as published; topic and difficulty added by this site.