Question Number 147689 by mathdanisur last updated on 22/Jul/21
$$\left(\mathrm{234}\right)_{\mathrm{5}} \:\centerdot\:\left(\mathrm{23}\right)_{\mathrm{5}} \:=\:\left({x}\right)_{\mathrm{5}} \:\:\Rightarrow\:\:\boldsymbol{{x}}=? \\ $$
Answered by Olaf_Thorendsen last updated on 22/Jul/21
$$\left(\mathrm{234}\right)_{\mathrm{5}} .\left(\mathrm{23}\right)_{\mathrm{5}} \\ $$$$=\:\left(\mathrm{2}.\mathrm{5}^{\mathrm{2}} +\mathrm{3}.\mathrm{5}^{\mathrm{1}} +\mathrm{4}.\mathrm{5}^{\mathrm{0}} \right)_{\mathrm{10}} .\left(\mathrm{2}.\mathrm{5}^{\mathrm{1}} +\mathrm{3}.\mathrm{5}^{\mathrm{0}} \right)_{\mathrm{10}} \\ $$$$=\:\left(\mathrm{69}\right)_{\mathrm{10}} .\left(\mathrm{13}\right)_{\mathrm{10}} \:=\:\left(\mathrm{69}×\mathrm{13}\right)_{\mathrm{10}} \:=\:\left(\mathrm{897}\right)_{\mathrm{10}} \\ $$$$=\:\left(\mathrm{625}+\mathrm{2}×\mathrm{125}+\mathrm{4}×\mathrm{5}+\mathrm{2}\right)_{\mathrm{10}} \\ $$$$=\:\left(\mathrm{1}.\mathrm{5}^{\mathrm{4}} +\mathrm{2}.\mathrm{5}^{\mathrm{3}} +\mathrm{4}.\mathrm{5}^{\mathrm{1}} +\mathrm{2}.\mathrm{5}^{\mathrm{0}} \right)_{\mathrm{10}} \\ $$$$=\:\left(\mathrm{12042}\right)_{\mathrm{5}} \\ $$$${x}\:=\:\mathrm{12042} \\ $$
Commented by mathdanisur last updated on 22/Jul/21
$$\left.{T}\left.{hank}\:{you}\:{Ser},\:{but}\:\uparrow\:\right)_{\mathrm{10}\:?} \:{or}\:\right)_{\mathrm{5}\:?} \\ $$
Commented by Olaf_Thorendsen last updated on 22/Jul/21
$$\left(\mathrm{897}\right)_{\mathrm{10}} \:=\:\left(\mathrm{12042}\right)_{\mathrm{5}} \\ $$
Commented by mathdanisur last updated on 23/Jul/21
$${thankyou}\:{Sir} \\ $$