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Question Number 19799 by Tinkutara last updated on 15/Aug/17
Foranaturalnumberb,letN(b)denotethenumberofnaturalnumbersaforwhichtheequationx2+ax+b=0hasintegerroots.WhatisthesmallestvalueofbforwhichN(b)=20?
Commented by dioph last updated on 16/Aug/17
x=−a±a2−4b2x1+x2=−ax1x2=bN(b)=n2,wherenisthenumberofdivisorsofbn=40⇒b=p1n1...pknk(n1+1)...(nk+1)=40Thesmallestnumberwith40naturaldivisorsIcanfindisb=24.3.5.7=1680
Commented by Tinkutara last updated on 16/Aug/17
ThankyouverymuchSir!
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