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41-47-find-remaider-

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

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