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Question Number 133764 by liberty last updated on 24/Feb/21

Answered by EDWIN88 last updated on 24/Feb/21

let x be a page number was counted twice Assuming the pages start counting at 1 and count continously up we use formula (1+2+3+...+n)+x = 1999 ⇒x +((n(n+1))/2) = 1999 ; since x is positive integer number ⇒we get n = 62 , so x = 1999−((62×63)/2) x = 1999−1953 = 46

$$\:\mathrm{let}\:\mathrm{x}\:\mathrm{be}\:\mathrm{a}\:\mathrm{page}\:\mathrm{number}\:\mathrm{was}\:\mathrm{counted}\:\mathrm{twice} \\ $$$$\mathrm{Assuming}\:\mathrm{the}\:\mathrm{pages}\:\mathrm{start}\:\mathrm{counting}\:\mathrm{at}\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{count}\:\mathrm{continously}\:\mathrm{up}\:\mathrm{we}\:\mathrm{use}\:\mathrm{formula} \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{n}\right)+\mathrm{x}\:=\:\mathrm{1999}\: \\ $$$$\Rightarrow\mathrm{x}\:+\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:=\:\mathrm{1999}\:;\:\mathrm{since}\:\mathrm{x}\:\mathrm{is}\:\mathrm{positive}\: \\ $$$$\mathrm{integer}\:\mathrm{number} \\ $$$$\Rightarrow\mathrm{we}\:\mathrm{get}\:\mathrm{n}\:=\:\mathrm{62}\:,\:\mathrm{so}\:\mathrm{x}\:=\:\mathrm{1999}−\frac{\mathrm{62}×\mathrm{63}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:\mathrm{1999}−\mathrm{1953}\:=\:\mathrm{46} \\ $$

Answered by mr W last updated on 24/Feb/21

say the book has totally n pages, the page number x is counted twice. 1≤x≤n ((n(n+1))/2)+x=1999 1999=((n(n+1))/2)+x≥((n(n+1))/2)+1 ⇒n^2 +n−3996≤0 −62.7≤n≤62.7 1999=((n(n+1))/2)+x≤((n(n+1))/2)+n ⇒n^2 +3n−3998≥0 n≤−64.7 or n≥61.7 ⇒n=62 ⇒x=1999−((62×63)/2)=46

$${say}\:{the}\:{book}\:{has}\:{totally}\:{n}\:{pages},\:{the} \\ $$$${page}\:{number}\:{x}\:{is}\:{counted}\:{twice}. \\ $$$$\mathrm{1}\leqslant{x}\leqslant{n} \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{x}=\mathrm{1999} \\ $$$$\mathrm{1999}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{x}\geqslant\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\mathrm{1} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +{n}−\mathrm{3996}\leqslant\mathrm{0} \\ $$$$−\mathrm{62}.\mathrm{7}\leqslant{n}\leqslant\mathrm{62}.\mathrm{7} \\ $$$$\mathrm{1999}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{x}\leqslant\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{n} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{3998}\geqslant\mathrm{0} \\ $$$${n}\leqslant−\mathrm{64}.\mathrm{7}\:{or}\:{n}\geqslant\mathrm{61}.\mathrm{7} \\ $$$$\Rightarrow{n}=\mathrm{62} \\ $$$$\Rightarrow{x}=\mathrm{1999}−\frac{\mathrm{62}×\mathrm{63}}{\mathrm{2}}=\mathrm{46} \\ $$

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