Question Number 213956 by Ari last updated on 22/Nov/24
Answered by A5T last updated on 22/Nov/24
$${Let}\:{the}\:{numbers}\:{be}:\:\left({x}−\mathrm{2},{x}−\mathrm{1},{x},{x}+\mathrm{1},{x}+\mathrm{2}\right) \\ $$$${x}=\mathrm{5}{a}+\mathrm{2}=\mathrm{7}{b}+\mathrm{1}=\mathrm{9}{c}=\mathrm{11}{d}−\mathrm{1}=\mathrm{13}{e}−\mathrm{2} \\ $$$$\mathrm{7}{b}+\mathrm{1}\equiv\mathrm{2}\left({mod}\:\mathrm{5}\right)\Rightarrow{b}\equiv\mathrm{3}\left({mod}\:\mathrm{5}\right)\Rightarrow{b}=\mathrm{5}{f}+\mathrm{3} \\ $$$$\Rightarrow\mathrm{7}\left(\mathrm{5}{f}+\mathrm{3}\right)+\mathrm{1}\equiv\mathrm{0}\left({mod}\:\mathrm{9}\right)\Rightarrow{f}\equiv\mathrm{4}\left({mod}\:\mathrm{9}\right)\Rightarrow{f}=\mathrm{9}{g}+\mathrm{4} \\ $$$$\Rightarrow\mathrm{7}{g}\equiv\mathrm{2}\left({mod}\:\mathrm{11}\right)\Rightarrow{g}\equiv\mathrm{5}\left({mod}\:\mathrm{11}\right)\Rightarrow{g}=\mathrm{11}{h}+\mathrm{5} \\ $$$$\mathrm{7}\left(\mathrm{495}{h}+\mathrm{248}\right)+\mathrm{1}\equiv\mathrm{11}\left({mod}\:\mathrm{13}\right) \\ $$$${h}\equiv\mathrm{6}\left({mod}\:\mathrm{13}\right);\:{h}=\mathrm{13}{i}+\mathrm{6} \\ $$$$\Rightarrow{x}=\mathrm{3465}\left(\mathrm{13}{i}+\mathrm{6}\right)+\mathrm{1737}=\mathrm{45045}{i}+\mathrm{22527} \\ $$$${There}\:{are}\:{infinitely}\:{many}\:{solutions}\: \\ $$$${For}\:{example}:\:\left(\mathrm{22525},\mathrm{22526},\mathrm{22527},\mathrm{22528},\mathrm{22529}\right) \\ $$
Commented by A5T last updated on 23/Nov/24
$${Yea},\:{for}\:{five}-{digit}\:{numbers}. \\ $$
Commented by nikif99 last updated on 23/Nov/24
$${if}\:{i}=\mathrm{0}\Rightarrow{x}=\mathrm{22527}\:\checkmark \\ $$$${if}\:{i}=\mathrm{1}\Rightarrow{x}=\mathrm{67572}\:\checkmark \\ $$$${if}\:{i}=\mathrm{2}\Rightarrow{x}>\mathrm{99999} \\ $$
Answered by TonyCWX08 last updated on 11/Jan/25
$${x}\equiv\mathrm{0}\left({mod}\mathrm{5}\right) \\ $$$${x}+\mathrm{1}\equiv\mathrm{0}\left({mod}\mathrm{7}\right)\Rightarrow{x}\equiv\mathrm{6}\left({mod}\mathrm{7}\right) \\ $$$${x}+\mathrm{2}\equiv\mathrm{0}\left({mod}\mathrm{9}\right)\Rightarrow{x}\equiv\mathrm{7}\left({mod}\mathrm{9}\right) \\ $$$${x}+\mathrm{3}\equiv\mathrm{0}\left({mod}\mathrm{11}\right)\Rightarrow{x}\equiv\mathrm{8}\left({mod}\mathrm{11}\right) \\ $$$${x}+\mathrm{4}\equiv\mathrm{0}\left({mod}\mathrm{13}\right)\Rightarrow{x}\equiv\mathrm{9}\left({mod}\mathrm{13}\right) \\ $$$$ \\ $$$${M}=\mathrm{7}×\mathrm{9}×\mathrm{11}×\mathrm{13}=\mathrm{9009} \\ $$$$ \\ $$$${a}_{{i}} \ast\frac{{M}}{{m}_{{i}} }\ast\left(\left(\frac{{M}}{{m}_{{i}} }\right)^{−\mathrm{1}} \left({mod}\:{m}_{{i}} \right)\right) \\ $$$$ \\ $$$${First}\:{Term} \\ $$$$=\mathrm{6}×\frac{\mathrm{9009}}{\mathrm{7}}×\left(\left(\frac{\mathrm{9009}}{\mathrm{7}}\right)^{−\mathrm{1}} \left({mod}\mathrm{7}\right)\right) \\ $$$$=\mathrm{6}×\mathrm{1287}×\mathrm{6} \\ $$$$=\mathrm{46332} \\ $$$$ \\ $$$${Second}\:{Term} \\ $$$$=\mathrm{7}×\mathrm{1001}×\mathrm{7} \\ $$$$=\mathrm{49049} \\ $$$$ \\ $$$${Third}\:{Term}\:=\:\mathrm{52416} \\ $$$$ \\ $$$${Fourth}\:{Term}\:=\:\mathrm{56133} \\ $$$$ \\ $$$${x}\equiv\left(\mathrm{46332}+\mathrm{49049}+\mathrm{52416}+\mathrm{56133}\right)\left({mod}\mathrm{9009}\right) \\ $$$${x}\equiv\mathrm{5732}\left({mod}\mathrm{9009}\right) \\ $$$${x}=\mathrm{9009}{n}+\mathrm{5732},\:{n}\in\mathbb{Z} \\ $$