Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

Question Number 117369 by bemath last updated on 11/Oct/20

Find the units digit of 2013^1 +2013^2 +2013^3 +...+2013^(2013)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{units}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{2013}^{\mathrm{1}} +\mathrm{2013}^{\mathrm{2}} +\mathrm{2013}^{\mathrm{3}} +...+\mathrm{2013}^{\mathrm{2013}} \\ $$

Answered by JDamian last updated on 11/Oct/20

S=2013^1 +2013^2 + ∙∙∙ + 2013^(2013) S mod 10 = (3^1 +3^2 +∙∙∙+3^(2013) )mod 10 The last digit of these powers are 3, 9, 7, 1, 3, 9, 7, 1, ∙∙∙ And the last digit of the sum 3, 2, 9, 0, 3, 2, 9, 0, ∙∙∙ As 2013 ≡ 1 mod 4 the answer is 3

$${S}=\mathrm{2013}^{\mathrm{1}} +\mathrm{2013}^{\mathrm{2}} +\:\centerdot\centerdot\centerdot\:+\:\mathrm{2013}^{\mathrm{2013}} \\ $$$${S}\:{mod}\:\mathrm{10}\:=\:\left(\mathrm{3}^{\mathrm{1}} +\mathrm{3}^{\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{3}^{\mathrm{2013}} \right){mod}\:\mathrm{10} \\ $$$$\mathrm{The}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{these}\:\mathrm{powers}\:\mathrm{are} \\ $$$$\mathrm{3},\:\mathrm{9},\:\mathrm{7},\:\mathrm{1},\:\mathrm{3},\:\mathrm{9},\:\mathrm{7},\:\mathrm{1},\:\centerdot\centerdot\centerdot \\ $$$$\mathrm{And}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{3},\:\mathrm{2},\:\mathrm{9},\:\mathrm{0},\:\mathrm{3},\:\mathrm{2},\:\mathrm{9},\:\mathrm{0},\:\centerdot\centerdot\centerdot \\ $$$$\mathrm{As}\:\:\:\mathrm{2013}\:\equiv\:\mathrm{1}\:{mod}\:\mathrm{4}\:\:\:\:{the}\:{answer}\:{is}\:\mathrm{3} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com