Question Number 85183 by Serlea last updated on 19/Mar/20
$$\mathrm{Hi}\:\mathrm{veterans} \\ $$$$\mathrm{Serlea}\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{three}\:\mathrm{digits}\:\mathrm{of}: \\ $$$$\mathrm{3005}^{\mathrm{11}} +\mathrm{3005}^{\mathrm{12}} +\mathrm{3005}^{\mathrm{13}} +...+\mathrm{3005}^{\mathrm{3002}} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 19/Mar/20
$${let}\:{say}\:{A}\overset{\mathrm{3}} {=}{B}\:{means}\:``{the}\:{last}\:{three} \\ $$$${digits}\:{from}\:{A}\:{are}\:{equal}\:{to}\:{the}\:{last} \\ $$$${three}\:{digits}\:{from}\:{B}''. \\ $$$$\mathrm{3005}^{{n}} \overset{\mathrm{3}} {=}\mathrm{5}^{{n}} \\ $$$$\mathrm{5}^{{odd}} \overset{\mathrm{3}} {=}\mathrm{125}\:\:\left({with}\:{odd}\geqslant\mathrm{3}\right) \\ $$$$\mathrm{5}^{{even}} \overset{\mathrm{3}} {=}\mathrm{625}\:\left({with}\:{even}\geqslant\mathrm{4}\right) \\ $$$$ \\ $$$$\mathrm{3005}^{\mathrm{11}} +\mathrm{3005}^{\mathrm{12}} +\mathrm{3005}^{\mathrm{13}} +...+\mathrm{3005}^{\mathrm{3002}} \\ $$$$\overset{\mathrm{3}} {=}\mathrm{5}^{\mathrm{11}} +\mathrm{5}^{\mathrm{12}} +\mathrm{5}^{\mathrm{13}} +\mathrm{5}^{\mathrm{14}} +...+\mathrm{5}^{\mathrm{3001}} +\mathrm{5}^{\mathrm{3002}} \\ $$$$\overset{\mathrm{3}} {=}\left(\mathrm{125}+\mathrm{625}\right)+\left(\mathrm{125}+\mathrm{625}\right)+...+\left(\mathrm{125}+\mathrm{625}\right) \\ $$$$\overset{\mathrm{3}} {=}\mathrm{750}×\mathrm{1496} \\ $$$$\overset{\mathrm{3}} {=}\mathrm{000}\:\Rightarrow{answer} \\ $$