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Question Number 19700 by Tinkutara last updated on 14/Aug/17

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{possible}\mathrm{value}\mathrm{of}$$$$k\mathrm{for}\mathrm{which}2013\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{a}$$$$\mathrm{sum}\mathrm{of}k\mathrm{consecutive}\mathrm{positive}\mathrm{integers}?$$

Answered by mrW1 last updated on 15/Aug/17

$$\mathrm{keep}\mathrm{in}\mathrm{mind}:2013=3\times 11\times 61$$$$\mathrm{let}\mathrm{a}=\mathrm{the}\mathrm{first}\mathrm{number},\mathrm{a}⩾1$$$$$$$$\mathrm{if}\mathrm{k}\mathrm{is}\mathrm{even},\mathrm{i}.\mathrm{e}.\mathrm{k}=2\mathrm{n}$$$$\mathrm{a}=\frac{2013}{2\mathrm{n}}-\frac{2\mathrm{n}-1}{2}=\frac{1}{2}(\frac{2013}{\mathrm{n}}-2\mathrm{n}+1)⩾1$$$$\Rightarrow \frac{2013}{\mathrm{n}}-2\mathrm{n}⩾1$$$$\Rightarrow {2\mathrm{n}}^2+\mathrm{n}-2013⩽0$$$$\mathrm{n}⩽\frac{-1+\sqrt{1+8\times 2013}}{4}\approx 31.5$$$$\mathrm{since}2013=1\times 3\times 11\times 61$$$$\Rightarrow \mathrm{n}=1,3,11$$$$\Rightarrow \mathrm{k}=2,6,22$$$$$$$$\mathrm{if}\mathrm{k}\mathrm{is}\mathrm{odd},\mathrm{i}.\mathrm{e}.\mathrm{k}=2\mathrm{n}+1$$$$\mathrm{a}=\frac{2013}{2\mathrm{n}+1}-\mathrm{n}⩾1$$$$2013-{2\mathrm{n}}^2-\mathrm{n}⩾2\mathrm{n}+1$$$${2\mathrm{n}}^2+3\mathrm{n}-2012⩽0$$$$\mathrm{n}⩽\frac{-3+\sqrt{9+8\times 2012}}{4}\approx 30.9$$$$\Rightarrow 2\mathrm{n}+1⩽62.9$$$$\mathrm{since}2013=1\times 3\times 11\times 61$$$$\Rightarrow \mathrm{k}=2\mathrm{n}+1=1,3,11,33,61$$$$$$$$\mathrm{all}\mathrm{possibilities}\mathrm{are}$$$$\mathrm{k}=1,2,3,6,11,22,33,61$$$${\mathrm{k}}_{\max }=61$$

Commented by Tinkutara last updated on 15/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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