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Lê Anh Duy

11/03/2018 at 12:58
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Given \(A=2018^2-2017^2+2016^2-2015^2+...+2^2-1^2\)

Find the last two digits of A


Grade 7


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    FC Alan Walker 11/03/2018 at 13:54

    We have: \(A=2018^2-2017^2+2016^2-2015^2+...+2^2-1^2\)  

                        \(=\left(2018-2017\right)\left(2018+2017\right)+\left(2016-2015\right)\left(2016+2015\right)+...+\left(2-1\right)\left(2+1\right)\)

                        \(=2018+2017+2016+2015+...+2+1\)

                        \(=\dfrac{2018.2019}{2}=2037171\).

    So the last two digits of A is 71.

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