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Question Number 115396 by Sudip last updated on 25/Sep/20

If a page is torn from the middle of a book, then the sum of the remaining pages is 718797 so what is the number of torn pages?

$$\mathrm{If}\:\mathrm{a}\:\mathrm{page}\:\mathrm{is}\:\mathrm{torn}\:\mathrm{from}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{book},\:\mathrm{then} \\ $$$$\mathrm{the}\:\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{pages}\:\mathrm{is}\:\mathrm{718797}\:\mathrm{so} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{torn}\:\mathrm{pages}? \\ $$

Answered by PRITHWISH SEN 2 last updated on 25/Sep/20

901 & 902

$$\mathrm{901}\:\&\:\mathrm{902} \\ $$

Answered by TANMAY PANACEA last updated on 25/Sep/20

S=(n/2)[2×1+(n−1)1] S=718797+x+x+1=((n^2 +n)/2) n^2 +n=2(2x+718798) wait...

$${S}=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}×\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{1}\right] \\ $$$${S}=\mathrm{718797}+{x}+{x}+\mathrm{1}=\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} +{n}=\mathrm{2}\left(\mathrm{2}{x}+\mathrm{718798}\right) \\ $$$${wait}... \\ $$

Answered by mr W last updated on 26/Sep/20

say the book has totally n pages. one sheet with page number m and m+1 is torn. ((n(n+1))/2)−(2m+1)=718797 n^2 +n−4(m+359399)=0 n=((−1+(√(1+16(m+281×1279))))/2) n>((−1+(√(1+16×281×1279)))/2)>1198 Δ=1+16(m+281×1279)=(2k+1)^2 ⇒n=k 4(m+281×1279)=k(k+1) ⇒k=4h ⇒m+281×1279=h(4h+1) m<n=k=4h m=h(4h+1)−281×1279<4h 4h^2 −3h−281×1279<0 ⇒h<((3+(√(9+16×281×1279)))/8)≈300.1 ⇒h≤300 4h=n>1198 ⇒h>((1198)/4)=299.5 ⇒h≥300 we see the only solution is h=300: n=4h=1200 m=h(4h+1)−281×1279=901 ⇒the book has 1200 pages, page 901 and 902 are torn.

$${say}\:{the}\:{book}\:{has}\:{totally}\:{n}\:{pages}. \\ $$$${one}\:{sheet}\:{with}\:{page}\:{number}\:{m}\:{and} \\ $$$${m}+\mathrm{1}\:{is}\:{torn}. \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\left(\mathrm{2}{m}+\mathrm{1}\right)=\mathrm{718797} \\ $$$${n}^{\mathrm{2}} +{n}−\mathrm{4}\left({m}+\mathrm{359399}\right)=\mathrm{0} \\ $$$${n}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{16}\left({m}+\mathrm{281}×\mathrm{1279}\right)}}{\mathrm{2}} \\ $$$${n}>\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{16}×\mathrm{281}×\mathrm{1279}}}{\mathrm{2}}>\mathrm{1198} \\ $$$$ \\ $$$$\Delta=\mathrm{1}+\mathrm{16}\left({m}+\mathrm{281}×\mathrm{1279}\right)=\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{n}={k} \\ $$$$ \\ $$$$\mathrm{4}\left({m}+\mathrm{281}×\mathrm{1279}\right)={k}\left({k}+\mathrm{1}\right) \\ $$$$\Rightarrow{k}=\mathrm{4}{h} \\ $$$$\Rightarrow{m}+\mathrm{281}×\mathrm{1279}={h}\left(\mathrm{4}{h}+\mathrm{1}\right) \\ $$$$ \\ $$$${m}<{n}={k}=\mathrm{4}{h} \\ $$$${m}={h}\left(\mathrm{4}{h}+\mathrm{1}\right)−\mathrm{281}×\mathrm{1279}<\mathrm{4}{h} \\ $$$$\mathrm{4}{h}^{\mathrm{2}} −\mathrm{3}{h}−\mathrm{281}×\mathrm{1279}<\mathrm{0} \\ $$$$\Rightarrow{h}<\frac{\mathrm{3}+\sqrt{\mathrm{9}+\mathrm{16}×\mathrm{281}×\mathrm{1279}}}{\mathrm{8}}\approx\mathrm{300}.\mathrm{1} \\ $$$$\Rightarrow{h}\leqslant\mathrm{300} \\ $$$$ \\ $$$$\mathrm{4}{h}={n}>\mathrm{1198} \\ $$$$\Rightarrow{h}>\frac{\mathrm{1198}}{\mathrm{4}}=\mathrm{299}.\mathrm{5} \\ $$$$\Rightarrow{h}\geqslant\mathrm{300} \\ $$$$ \\ $$$${we}\:{see}\:{the}\:{only}\:{solution}\:{is}\:{h}=\mathrm{300}: \\ $$$${n}=\mathrm{4}{h}=\mathrm{1200} \\ $$$${m}={h}\left(\mathrm{4}{h}+\mathrm{1}\right)−\mathrm{281}×\mathrm{1279}=\mathrm{901} \\ $$$$ \\ $$$$\Rightarrow{the}\:{book}\:{has}\:\mathrm{1200}\:{pages}, \\ $$$${page}\:\mathrm{901}\:{and}\:\mathrm{902}\:{are}\:{torn}. \\ $$

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