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Number theory Difficulty 5.8 AIME, harder Prove it

Theorem 2 Under the notation of Theorem 1, let
g~l=M0M01(1)hh(1)5hl(0)+M1M11g1h1(1)++MrMr1grhr(r),1lr,\begin{aligned} \widetilde{g}_{l}= & M_{0} M_{0}^{-1}(-1)^{h_{h}^{(-1)}} 5^{h_{l}^{(0)}}+M_{1} M_{1}^{-1} g_{1}^{h_{1}^{(1)}} \\ & +\cdots+M_{r} M_{r}^{-1} g_{r}^{h_{r}^{(r)}}, \quad-1 \leqslant l \leqslant r, \end{aligned}

where we take
hl(D)=1,hl(j)=0,jl,1jrh_{l}^{(D)}=1, \quad h_{l}^{(j)}=0, \quad j \neq l,-1 \leqslant j \leqslant r

Then
g~11(mod2α0),g~05(mod2a0)g~jgj(modpjαj),1jr,\begin{array}{c} \tilde{g}_{-1} \equiv-1\left(\bmod 2^{\alpha_{0}}\right), \quad \widetilde{g}_{0} \equiv 5\left(\bmod 2^{a_{0}}\right) \\ \widetilde{g}_{j} \equiv g_{j}\left(\bmod p_{j}^{\alpha_{j}}\right), \quad 1 \leqslant j \leqslant r, \end{array}

and the following φ(m)\varphi(m) numbers form a complete set of reduced residues modulo mm:
{x=g~1r1(1)g~0(0)g~1(1)g~r(r)0γ(j)<cj,1j<r\left\{\begin{array}{l} x=\widetilde{g}_{-1}^{r_{-1}^{(1)}} \widetilde{\mathrm{g}}_{0}^{(0)} \widetilde{\mathrm{g}}_{1}^{(1)} \cdots \widetilde{\mathrm{g}}_{r}^{(r)} \\ 0 \leqslant \gamma^{(j)}<c_{j},-1 \leqslant j<r \end{array}\right.

Solution

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.