Question Number 146224 by mathdanisur last updated on 12/Jul/21
Answered by gsk2684 last updated on 12/Jul/21
$$\mathrm{1}!\:\mathrm{2}!\:\mathrm{3}!\:...\:\left({x}−\mathrm{1}\right)!\:=\mathrm{288}=\mathrm{2}^{\mathrm{5}} \mathrm{3}^{\mathrm{2}} \\ $$$${x}−\mathrm{1}\leqslant\mathrm{4} \\ $$$${observing}\:\mathrm{1}!\:\mathrm{2}!\:\mathrm{3}!\:\mathrm{4}!=\mathrm{288} \\ $$$${x}=\mathrm{5} \\ $$
Commented by mathdanisur last updated on 12/Jul/21
$${cool}\:{Ser}\:{thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Jul/21
$$\:\mathrm{0}!.\mathrm{1}!.\mathrm{2}!.\mathrm{3}!....\left({x}−\mathrm{1}\right)!=\mathrm{288} \\ $$$$\Rightarrow\:\left({x}−\mathrm{1}\right)!\:\mid\:\mathrm{288}\:\wedge\:{x}!\:\nmid\:\mathrm{288}\Rightarrow{x}−\mathrm{1}=\mathrm{4} \\ $$$$\Rightarrow{x}=\mathrm{5} \\ $$
Commented by mathdanisur last updated on 12/Jul/21
$${Cool}\:{thanks}\:{Ser} \\ $$