Question Number 187869 by BaliramKumar last updated on 23/Feb/23
$$\frac{\mathrm{41}!}{\mathrm{47}}\:{find}\:{remaider} \\ $$
Answered by Rasheed.Sindhi last updated on 23/Feb/23
$${According}\:{to}\:{Wilson}'{s}\:{theorem}: \\ $$$$\left(\mathrm{47}−\mathrm{1}\right)!\equiv−\mathrm{1}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{46}!\equiv−\mathrm{1}+\mathrm{47}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{46}!\equiv\mathrm{46}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{45}!\equiv\mathrm{1}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{45}×\mathrm{44}!\equiv\mathrm{1}+\mathrm{47}×\mathrm{22}=\mathrm{1035}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{44}!\equiv\mathrm{23}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{44}×\mathrm{43}!\equiv\mathrm{23}+\mathrm{47}×\mathrm{7}=\mathrm{352}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{43}!\equiv\mathrm{8}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{43}×\mathrm{42}!\equiv\mathrm{8}+\mathrm{47}×\mathrm{41}=\mathrm{1935}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{42}!\equiv\mathrm{45}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{42}×\mathrm{41}!\equiv\mathrm{45}+\mathrm{47}×\mathrm{33}=\mathrm{1596}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{41}!\equiv\mathrm{38}\left({mod}\:\mathrm{47}\right) \\ $$
Answered by BaliramKumar last updated on 23/Feb/23
$$ \\ $$$${other}\:{approach} \\ $$$${According}\:{to}\:{Wilson}'{s}\:{theorem}: \\ $$$$\left(\mathrm{47}−\mathrm{1}\right)!\equiv−\mathrm{1}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{46}!\equiv−\mathrm{1}\:\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{46}×\mathrm{45}×\mathrm{44}×\mathrm{43}×\mathrm{42}×\mathrm{41}!\equiv−\mathrm{1}\left({mod}\:\mathrm{47}\right) \\ $$$$\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)\left(−\mathrm{4}\right)\left(−\mathrm{5}\right)\mathrm{41}!\equiv−\mathrm{1}\left({mod}\:\mathrm{47}\right) \\ $$$$−\mathrm{120}×\mathrm{41}!\equiv−\mathrm{1}\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{120}×\mathrm{41}!\:\equiv\:\mathrm{1}\:\:\:\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{120}×\mathrm{41}!\:\equiv\:\mathrm{48}\:\:\:\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{5}×\mathrm{41}!\:\equiv\:\mathrm{2}\:\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{5}×\mathrm{41}!\:\equiv\:\mathrm{2}+\mathrm{47}×\mathrm{4}\:\left({mod}\:\mathrm{47}\right) \\ $$$$\cancel{\mathrm{5}}×\mathrm{41}!\:\equiv\:\cancel{\mathrm{190}}\:\left({mod}\:\mathrm{47}\right) \\ $$$$\mathrm{41}!\equiv\mathrm{38}\left({mod}\:\mathrm{47}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 24/Feb/23
$$\cap\boldsymbol{\mathrm{i}}\subset\in! \\ $$