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41-3-42-3-43-3-59-3-Find-the-last-three-digits-of-the-number-




Question Number 147661 by mathdanisur last updated on 22/Jul/21
41^3  + 42^3  + 43^3  + ... + 59^3   Find the last three digits of the  number
$$\mathrm{41}^{\mathrm{3}} \:+\:\mathrm{42}^{\mathrm{3}} \:+\:\mathrm{43}^{\mathrm{3}} \:+\:…\:+\:\mathrm{59}^{\mathrm{3}} \\ $$$${Find}\:{the}\:{last}\:{three}\:{digits}\:{of}\:{the} \\ $$$${number} \\ $$
Answered by nimnim last updated on 22/Jul/21
S=(1^3 +2^3 +3^3 +....+59^3 )−(1^3 +2^3 +3^3 +....+40^3 )     =(((59×60)/2))^2 −(((40×41)/2))^2      =(1770)^2 −(820)^2      =(..70−..20)(..70+..20) → last two digits     =..50×..90     =....500→last three digits
$${S}=\left(\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +….+\mathrm{59}^{\mathrm{3}} \right)−\left(\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +….+\mathrm{40}^{\mathrm{3}} \right) \\ $$$$\:\:\:=\left(\frac{\mathrm{59}×\mathrm{60}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{40}×\mathrm{41}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:=\left(\cancel{\mathrm{17}70}\right)^{\mathrm{2}} −\left(\cancel{\mathrm{8}20}\right)^{\mathrm{2}} \\ $$$$\:\:\:=\left(..\mathrm{70}−..\mathrm{20}\right)\left(..\mathrm{70}+..\mathrm{20}\right)\:\rightarrow\:{last}\:{two}\:{digits} \\ $$$$\:\:\:=..\mathrm{50}×..\mathrm{90} \\ $$$$\:\:\:=….\mathrm{500}\rightarrow{last}\:{three}\:{digits} \\ $$
Commented by mathdanisur last updated on 23/Jul/21
Sir, (((59∙60)/2))−... can you explain this  part please.? how did you get it.?
$${Sir},\:\left(\frac{\mathrm{59}\centerdot\mathrm{60}}{\mathrm{2}}\right)−…\:{can}\:{you}\:{explain}\:{this} \\ $$$${part}\:{please}.?\:{how}\:{did}\:{you}\:{get}\:{it}.? \\ $$
Commented by mathdanisur last updated on 22/Jul/21
thanks Sir
$${thanks}\:{Sir} \\ $$
Commented by mathdanisur last updated on 23/Jul/21
(((59∙60)/2))^2 −...
$$\left(\frac{\mathrm{59}\centerdot\mathrm{60}}{\mathrm{2}}\right)^{\mathrm{2}} −… \\ $$

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