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Category: Algebra

Question-199381

Question Number 199381 by Mingma last updated on 02/Nov/23 Answered by witcher3 last updated on 02/Nov/23 (202)=2.101(201)!k0[202]0,k{1,201){2,101)J201!101+201!2[202]$$\frac{\mathrm{201}!}{\mathrm{2}}=\mathrm{202}.\mathrm{3}.\mathrm{2}.\underset{\mathrm{k}=\mathrm{5},\mathrm{k}\neq\mathrm{101}} {\overset{\mathrm{201}} {\prod}}\mathrm{k}\equiv\mathrm{0}\left[\mathrm{202}\right]…