All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 165831 by HongKing last updated on 09/Feb/22
Findtheintegerpartofthenumber:2015⋅2016⋅20173
Commented by mr W last updated on 09/Feb/22
x(x+1)(x+2)3>x×x×x3=xx(x+1)(x+2)3=(x+1)((x+1)2−1)3<(x+1)(x+1)23=x+1x<x(x+1)(x+2)3<x+1⇒[x(x+1)(x+2)3]=x⇒[2015×2016×20173=2015
Answered by naka3546 last updated on 09/Feb/22
Letx=20162015⋅2016⋅20173=(x−1)⋅x⋅(x+1)3=x(x2−1)3=x3−x3<x33Rememberthatx(x+1)−3x+1<x(x+1)(x−1)2<x(x+1)(x−1)⋅(x−1)23<(x−1)⋅x⋅(x+1)32015=(x−1)33<x⋅(x2−1)3<x33=2016Integerpartof2015⋅2016⋅20173is2015.
Commented by naka3546 last updated on 09/Feb/22
yes,sir.
Terms of Service
Privacy Policy
Contact: info@tinkutara.com