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What-is-the-maximum-possible-value-of-k-for-which-2013-can-be-written-as-a-sum-of-k-consecutive-positive-integers-

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Question Number 19700 by Tinkutara last updated on 14/Aug/17
What is the maximum possible value of k for which 2013 can be written as a sum of k consecutive positive integers?
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of} \\ $$$${k}\:\mathrm{for}\:\mathrm{which}\:\mathrm{2013}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{sum}\:\mathrm{of}\:{k}\:\mathrm{consecutive}\:\mathrm{positive}\:\mathrm{integers}? \\ $$
Answered by mrW1 last updated on 15/Aug/17
keep in mind: 2013=3×11×61 let a=the first number, a≥1 if k is even, i.e. k=2n a=((2013)/(2n))−((2n−1)/2)=(1/2)(((2013)/n)−2n+1)≥1 ⇒((2013)/n)−2n≥1 ⇒2n^2 +n−2013≤0 n≤((−1+(√(1+8×2013)))/4)≈31.5 since 2013=1×3×11×61 ⇒n=1,3,11 ⇒k=2,6,22 if k is odd, i.e. k=2n+1 a=((2013)/(2n+1))−n≥1 2013−2n^2 −n≥2n+1 2n^2 +3n−2012≤0 n≤((−3+(√(9+8×2012)))/4)≈30.9 ⇒ 2n+1≤62.9 since 2013=1×3×11×61 ⇒k=2n+1=1,3,11,33,61 all possibilities are k=1,2,3,6,11,22,33,61 k_(max) =61
$$\mathrm{keep}\:\mathrm{in}\:\mathrm{mind}:\:\mathrm{2013}=\mathrm{3}×\mathrm{11}×\mathrm{61} \\ $$$$\mathrm{let}\:\mathrm{a}=\mathrm{the}\:\mathrm{first}\:\mathrm{number},\:\mathrm{a}\geqslant\mathrm{1} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{k}\:\mathrm{is}\:\mathrm{even},\:\mathrm{i}.\mathrm{e}.\:\mathrm{k}=\mathrm{2n} \\ $$$$\mathrm{a}=\frac{\mathrm{2013}}{\mathrm{2n}}−\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2013}}{\mathrm{n}}−\mathrm{2n}+\mathrm{1}\right)\geqslant\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{2013}}{\mathrm{n}}−\mathrm{2n}\geqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{2n}^{\mathrm{2}} +\mathrm{n}−\mathrm{2013}\leqslant\mathrm{0} \\ $$$$\mathrm{n}\leqslant\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{8}×\mathrm{2013}}}{\mathrm{4}}\approx\mathrm{31}.\mathrm{5} \\ $$$$\mathrm{since}\:\mathrm{2013}=\mathrm{1}×\mathrm{3}×\mathrm{11}×\mathrm{61} \\ $$$$\Rightarrow\mathrm{n}=\mathrm{1},\mathrm{3},\mathrm{11} \\ $$$$\Rightarrow\mathrm{k}=\mathrm{2},\mathrm{6},\mathrm{22} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{k}\:\mathrm{is}\:\mathrm{odd},\:\mathrm{i}.\mathrm{e}.\:\mathrm{k}=\mathrm{2n}+\mathrm{1} \\ $$$$\mathrm{a}=\frac{\mathrm{2013}}{\mathrm{2n}+\mathrm{1}}−\mathrm{n}\geqslant\mathrm{1} \\ $$$$\mathrm{2013}−\mathrm{2n}^{\mathrm{2}} −\mathrm{n}\geqslant\mathrm{2n}+\mathrm{1} \\ $$$$\mathrm{2n}^{\mathrm{2}} +\mathrm{3n}−\mathrm{2012}\leqslant\mathrm{0} \\ $$$$\mathrm{n}\leqslant\frac{−\mathrm{3}+\sqrt{\mathrm{9}+\mathrm{8}×\mathrm{2012}}}{\mathrm{4}}\approx\mathrm{30}.\mathrm{9} \\ $$$$\Rightarrow\:\mathrm{2n}+\mathrm{1}\leqslant\mathrm{62}.\mathrm{9} \\ $$$$\mathrm{since}\:\mathrm{2013}=\mathrm{1}×\mathrm{3}×\mathrm{11}×\mathrm{61} \\ $$$$\Rightarrow\mathrm{k}=\mathrm{2n}+\mathrm{1}=\mathrm{1},\mathrm{3},\mathrm{11},\mathrm{33},\mathrm{61} \\ $$$$ \\ $$$$\mathrm{all}\:\mathrm{possibilities}\:\mathrm{are} \\ $$$$\mathrm{k}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{6},\mathrm{11},\mathrm{22},\mathrm{33},\mathrm{61} \\ $$$$\mathrm{k}_{\mathrm{max}} =\mathrm{61} \\ $$
Commented by Tinkutara last updated on 15/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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