Question

restul împărțirii cu rest 51 al ne n=1x2x3x.......x20+105​

Answer 1

$\displaystyle\bf\\20!+105 = \mathcal{M}_{51}+x,~x~\in\mathbb{N},~x<51.\\descompunem~numarul~51~in~factori~primi,~51=3\cdot17.\\20!~contine~in~mod~evident~factorii~3~si~17,~20!=1\cdot2\cdot3\cdot...\cdot17\cdot...\cdot20,\\\implies 20!~|~2\cdot17=51.\\deci~restul~numarului~20!+105~la~51~este~dat~de~restul~impartirii~lui~105~la\\51.\\20!+105=\mathcal{M}_{51}+105=\mathcal{M}_{51}+\mathcal{M}_{51}+3=\mathcal{M}_{51}+3.\\asadar,~restul~va~fi~3.$