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

Answered by EDWIN88 last updated on 24/Feb/21

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

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

$$saythebookhastotallynpages,the$$$$pagenumberxiscountedtwice.$$$$1⩽x⩽n$$$$\frac{n(n+1)}{2}+x=1999$$$$1999=\frac{n(n+1)}{2}+x⩾\frac{n(n+1)}{2}+1$$$$\Rightarrow n^2+n-3996⩽0$$$$-62.7⩽n⩽62.7$$$$1999=\frac{n(n+1)}{2}+x⩽\frac{n(n+1)}{2}+n$$$$\Rightarrow n^2+3n-3998⩾0$$$$n⩽-64.7orn⩾61.7$$$$\Rightarrow n=62$$$$\Rightarrow x=1999-\frac{62\times 63}{2}=46$$

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