Question Number 209719 by hardmath last updated on 19/Jul/24
$$\mathrm{If}: \\ $$$$\mathrm{7}^{\mathrm{243}} \:\:=\:\:\overline {\mathrm{a}…\mathrm{bc}} \\ $$$$\mathrm{Find}: \\ $$$$\mathrm{b}\centerdot\mathrm{c}\:=\:? \\ $$
Answered by MM42 last updated on 19/Jul/24
$$\mathrm{7}^{\mathrm{4}} \overset{\mathrm{100}} {\equiv}\mathrm{1}\Rightarrow\mathrm{7}^{\mathrm{240}} \overset{\mathrm{100}} {\equiv}\mathrm{1}\Rightarrow\mathrm{7}^{\mathrm{243}} \overset{\mathrm{100}} {\equiv}\mathrm{7}^{\mathrm{3}} \overset{\mathrm{100}} {\equiv}\mathrm{43} \\ $$$$\Rightarrow{bc}=\mathrm{12}\:\checkmark \\ $$$$ \\ $$
Commented by hardmath last updated on 19/Jul/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{but}\:\boldsymbol{\mathrm{a}}\:=\:? \\ $$
Commented by MM42 last updated on 19/Jul/24
$${I}\:{dont}\:{know} \\ $$
Commented by Frix last updated on 22/Jul/24
$$\mathrm{log}_{\mathrm{10}} \:\mathrm{7}^{\mathrm{243}} \:=\mathrm{243log}_{\mathrm{10}} \:\mathrm{7}\:\approx\mathrm{243}×.\mathrm{845098}\approx \\ $$$$\approx\mathrm{205}.\mathrm{359} \\ $$$$\mathrm{7}^{\mathrm{243}} \approx\mathrm{10}^{\mathrm{205}} ×\mathrm{10}^{.\mathrm{359}} \\ $$$$\mathrm{10}^{.\mathrm{359}} \approx\mathrm{2}.\mathrm{29} \\ $$$$\mathrm{7}^{\mathrm{243}} \approx\mathrm{2}.\mathrm{29}×\mathrm{10}^{\mathrm{205}} \\ $$$${a}=\mathrm{2} \\ $$