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Question Number 47316 by peter frank last updated on 08/Nov/18

Commented by math1967 last updated on 08/Nov/18

$$mancontains2shoesand2paper$$

Commented by MJS last updated on 08/Nov/18

$$\left(1\right)\mathrm{try}\mathrm{to}\mathrm{buy}\mathrm{only}\mathrm{the}\mathrm{right}\mathrm{shoe}\left(s_r\right);{\mathrm{you}}^{\prime }\mathrm{ll}\mathrm{find}\mathrm{you}$$$$\mathrm{have}\mathrm{to}\mathrm{buy}\mathrm{the}\mathrm{pair}\Rightarrow s_r=s_r+s_l=s_l.\mathrm{but}\mathrm{here}$$$$\mathrm{we}\mathrm{have}{s}_r{s}_l+s_r{s}_l+s_r{s}_l=30\Rightarrow s_r{s}_l=10\mathrm{so}\mathrm{this}$$$$\mathrm{might}\mathrm{get}\mathrm{hard}\mathrm{to}\mathrm{solve}$$$$\left(2\right)\mathrm{two}\mathrm{boys}\mathrm{will}\mathrm{cause}\mathrm{much}\mathrm{more}\mathrm{trouble}$$$$\mathrm{than}\mathrm{one}\mathrm{boy}\left(b\right),\mathrm{I}\mathrm{have}4\mathrm{children},3\mathrm{boys}$$$$\mathrm{and}\mathrm{one}\mathrm{girl},\mathrm{together}\mathrm{they}\mathrm{caused}\approx 11.4066\mathrm{units}$$$$\mathrm{of}\mathrm{trouble},\mathrm{the}\mathrm{exact}\mathrm{formula}\mathrm{is}trouble=\frac{{\pi }^{boys}}{{\mathrm{e}}^{girls}},\mathrm{in}$$$$\mathrm{our}\mathrm{case}{\pi }^2\approx 9.86960\Rightarrow b+b=9.86960b\Rightarrow$$$$9.86960b+s_r{s}_l=20$$$$\left(3\right)\mathrm{buying}\mathrm{two}\mathrm{cones}\mathrm{you}\mathrm{will}\mathrm{probably}\mathrm{get}\mathrm{\%},$$$$\mathrm{so}c+c<2c\mathrm{and}\mathrm{probably}c+c<c^2\Rightarrow c>2$$$$c^2+c^2+b=13$$$$\left(4\right)\mathrm{the}\mathrm{boy}\mathrm{now}\mathrm{wears}\mathrm{shoes},\mathrm{they}\mathrm{rather}$$$$\mathrm{look}\mathrm{multiplicated}\mathrm{again},\mathrm{and}\mathrm{to}\mathrm{get}\mathrm{the}{\mathrm{boy}}^{\prime }\mathrm{s}$$$$\mathrm{feet}\mathrm{out}\mathrm{of}\mathrm{the}\mathrm{shoes}\mathrm{we}\mathrm{will}\mathrm{have}\mathrm{to}\mathrm{extract}$$$$\mathrm{the}\mathrm{root}\Rightarrow \sqrt[b]{s_r{s}_l}.\mathrm{a}\mathrm{boy}\mathrm{looking}\mathrm{like}\mathrm{this}\mathrm{one}$$$$\mathrm{wearing}\mathrm{pink}\mathrm{shoes}\mathrm{only}\mathrm{exists}\mathrm{in}{\mathrm{somebody}}^{\prime }\mathrm{s}$$$$\mathrm{weird}\mathrm{dreams}\Rightarrow \mathrm{i}\sqrt[b]{s_r{s}_l}\mathrm{and}\mathrm{he}\mathrm{holds}2\mathrm{cones},$$$$s_r+\mathrm{i}\sqrt[b]{s_r{s}_l}\times c^3=?$$$$\mathrm{this}\mathrm{cannot}\mathrm{be}\mathrm{solved}\mathrm{using}\mathrm{elementary}$$$$\mathrm{calculus}\mathrm{but}\mathrm{using}\mathrm{the}\mathrm{secret}\mathrm{formula}\mathrm{of}$$$$\mathrm{Ramanjuan}\mathrm{\&}\mathrm{Ostrogradski}\mathrm{at}\mathrm{least}\mathrm{we}\mathrm{get}$$$$\mathrm{the}\mathrm{Shoemaker}-\mathrm{Conestant}{\xi }_0=43$$

Commented by behi83417@gmail.com last updated on 08/Nov/18

$$6\left(shoes\right)=30\Rightarrow shoe=5$$$$2\left(man\right)+2\left(shoe\right)=20\Rightarrow man=5$$$$4\left(paper\right)+man=13\Rightarrow paper=2$$$$now:$$$$\left(shoe\right)+(man+2shoes+2paper)\times paper=$$$$=5+(5+2\times 5+2\times 2)\times 2=43.$$

Answered by ajfour last updated on 08/Nov/18

$$\frac{30+20}{2}-13.$$

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