Question Number 174144 by Shrinava last updated on 25/Jul/22
$$\mathrm{If}\:\:\mathrm{a},\mathrm{b}\in\mathbb{N}\:\:\mathrm{and}\:\:\mathrm{a}\:\centerdot\:\mathrm{b}\:=\:\mathrm{100} \\ $$$$\mathrm{a}^{\mathrm{6}} \:\centerdot\:\mathrm{b}^{\mathrm{8}} \:\:\mathrm{how}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{can}\:\mathrm{the}\:\mathrm{product} \\ $$$$\mathrm{have}\:\mathrm{at}\:\mathrm{most}? \\ $$$$\mathrm{Answer}:\:\:\mathrm{17} \\ $$
Answered by mr W last updated on 26/Jul/22
$${if}\:{a}\:{natural}\:{number}\:{N}\:{has}\:{n}\:{digits},\:{it} \\ $$$${means} \\ $$$$\mathrm{10}^{{n}−\mathrm{1}} \leqslant{N}<\mathrm{10}^{{n}} \\ $$$$\mathrm{log}\:\left(\mathrm{10}^{{n}−\mathrm{1}} \right)\leqslant\mathrm{log}\:\left({N}\right)<\mathrm{log}\:\mathrm{10}^{{n}} \\ $$$${n}−\mathrm{1}\leqslant\mathrm{log}\:\left({N}\right)<{n} \\ $$$$\mathrm{log}\:\left({N}\right)<{n}\leqslant\mathrm{log}\:\left({N}\right)+\mathrm{1} \\ $$$$\Rightarrow{n}=\lfloor\mathrm{log}\:\left({N}\right)\rfloor+\mathrm{1} \\ $$$${N}={a}^{\mathrm{6}} {b}^{\mathrm{8}} =\left(\frac{\mathrm{100}}{{b}}\right)^{\mathrm{6}} {b}^{\mathrm{8}} =\mathrm{10}^{\mathrm{12}} {b}^{\mathrm{2}} \\ $$$${N}_{{max}} =\mathrm{10}^{\mathrm{12}} ×\left(\mathrm{100}\right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{16}} \\ $$$${n}=\lfloor\mathrm{log}\:{N}_{{max}} \rfloor+\mathrm{1} \\ $$$$\:\:\:=\lfloor\mathrm{log}\:\mathrm{10}^{\mathrm{16}} \rfloor+\mathrm{1} \\ $$$$\:\:\:=\lfloor\mathrm{16}\rfloor+\mathrm{1} \\ $$$$\:\:\:=\mathrm{17} \\ $$
Commented by Shrinava last updated on 06/Aug/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$