Menu Close

Find-the-number-of-positive-integers-x-such-that-x-m-1-x-m-1-for-a-particular-integer-m-2-means-G-I-F-

Tinku Tara 0 Comments
Pin




Question Number 28186 by Tinkutara last updated on 25/Jan/18
Find the number of positive integers x such that [(x/(m−1))]=[(x/(m+1))], for a particular integer m≥2. [ ] means G.I.F.
Findthenumberofpositiveintegersxsuchthat[xm−1]=[xm+1],foraparticularintegerm⩾2.[]meansG.I.F.
Commented by abdo imad last updated on 25/Jan/18
if 0<x<m−1⇒0< (x/(m−1))<1 and 0<x<m+1 ⇒ [(x/(m−1))]=[(x/(m+1))]=0 let suppose x ≥m−1≥1 x=q(m−1) +r and q ,r integers 0≤r<m−1 (x/(m−1))=q +(r/(m−1)) ⇒[(x/(m−1))]=q+[(r/(m−1))]=q+0=q (x/(m+1))=((q(m−1)+r)/(m+1)) = ((q(m+1)−2q +r)/(m+1))=q + ((r−2q)/(m+1)) and [(x/(m+1))]=q +[((r−2q)/(m+1))] e ⇔q=q +[((r−2q)/(m+1))]⇔ [((r−2q)/(m+1))]=0 ⇔0≤ ((r−2q)/(m+1))<1 ⇔0≤ r−2q<m+1 but x = qm−q +r=q(m+1)+r−2q ⇒r−2q=x −q(m+1)⇒0≤x −q(m+1)<m+1 ⇒q(m+1)≤x<(q+1)(m+1)....
if0<x<m−1⇒0<xm−1<1and0<x<m+1⇒[xm−1]=[xm+1]=0letsupposex⩾m−1⩾1x=q(m−1)+randq,rintegers0⩽r<m−1xm−1=q+rm−1⇒[xm−1]=q+[rm−1]=q+0=qxm+1=q(m−1)+rm+1=q(m+1)−2q+rm+1=q+r−2qm+1and[xm+1]=q+[r−2qm+1]e⇔q=q+[r−2qm+1]⇔[r−2qm+1]=0⇔0⩽r−2qm+1<1⇔0⩽r−2q<m+1butx=qm−q+r=q(m+1)+r−2q⇒r−2q=x−q(m+1)⇒0⩽x−q(m+1)<m+1⇒q(m+1)⩽x<(q+1)(m+1)….

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *