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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
man contains 2shoes and 2paper
$${man}\:\:{contains}\:\mathrm{2}{shoes}\:{and}\:\mathrm{2}{paper} \\ $$
Commented by MJS last updated on 08/Nov/18
(1) try to buy only the right shoe (s_r ); you′ll find you have to buy the pair ⇒ s_r =s_r +s_l =s_l . but here we have s_r s_l +s_r s_l +s_r s_l =30 ⇒ s_r s_l =10 so this might get hard to solve (2) two boys will cause much more trouble than one boy (b), I have 4 children, 3 boys and one girl, together they caused ≈11.4066 units of trouble, the exact formula is trouble=(π^(boys) /e^(girls) ), in our case π^2 ≈9.86960 ⇒ b+b=9.86960b ⇒ 9.86960b+s_r s_l =20 (3) buying two cones you will probably get %, so c+c<2c and probably c+c<c^2 ⇒ c>2 c^2 +c^2 +b=13 (4) the boy now wears shoes, they rather look multiplicated again, and to get the boy′s feet out of the shoes we will have to extract the root ⇒ ((s_r s_l ))^(1/b) . a boy looking like this one wearing pink shoes only exists in somebody′s weird dreams ⇒ i((s_r s_l ))^(1/b) and he holds 2 cones, s_r +i((s_r s_l ))^(1/b) ×c^3 =? this cannot be solved using elementary calculus but using the secret formula of Ramanjuan&Ostrogradski at least we get the Shoemaker−Conestant ξ_0 =43
$$\left(\mathrm{1}\right)\:\mathrm{try}\:\mathrm{to}\:\mathrm{buy}\:\mathrm{only}\:\mathrm{the}\:\mathrm{right}\:\mathrm{shoe}\:\left({s}_{{r}} \right);\:\mathrm{you}'\mathrm{ll}\:\mathrm{find}\:\mathrm{you} \\ $$$$\mathrm{have}\:\mathrm{to}\:\mathrm{buy}\:\mathrm{the}\:\mathrm{pair}\:\Rightarrow\:{s}_{{r}} ={s}_{{r}} +{s}_{{l}} ={s}_{{l}} .\:\mathrm{but}\:\mathrm{here} \\ $$$$\mathrm{we}\:\mathrm{have}\:{s}_{{r}} {s}_{{l}} +{s}_{{r}} {s}_{{l}} +{s}_{{r}} {s}_{{l}} =\mathrm{30}\:\Rightarrow\:{s}_{{r}} {s}_{{l}} =\mathrm{10}\:\mathrm{so}\:\mathrm{this} \\ $$$$\mathrm{might}\:\mathrm{get}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{two}\:\mathrm{boys}\:\mathrm{will}\:\mathrm{cause}\:\mathrm{much}\:\mathrm{more}\:\mathrm{trouble} \\ $$$$\mathrm{than}\:\mathrm{one}\:\mathrm{boy}\:\left({b}\right),\:\mathrm{I}\:\mathrm{have}\:\mathrm{4}\:\mathrm{children},\:\mathrm{3}\:\mathrm{boys} \\ $$$$\mathrm{and}\:\mathrm{one}\:\mathrm{girl},\:\mathrm{together}\:\mathrm{they}\:\mathrm{caused}\:\approx\mathrm{11}.\mathrm{4066}\:\mathrm{units} \\ $$$$\mathrm{of}\:\mathrm{trouble},\:\mathrm{the}\:\mathrm{exact}\:\mathrm{formula}\:\mathrm{is}\:{trouble}=\frac{\pi^{{boys}} }{\mathrm{e}^{{girls}} },\:\mathrm{in} \\ $$$$\mathrm{our}\:\mathrm{case}\:\pi^{\mathrm{2}} \approx\mathrm{9}.\mathrm{86960}\:\Rightarrow\:{b}+{b}=\mathrm{9}.\mathrm{86960}{b}\:\Rightarrow \\ $$$$\mathrm{9}.\mathrm{86960}{b}+{s}_{{r}} {s}_{{l}} =\mathrm{20} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{buying}\:\mathrm{two}\:\mathrm{cones}\:\mathrm{you}\:\mathrm{will}\:\mathrm{probably}\:\mathrm{get}\:\%, \\ $$$$\mathrm{so}\:{c}+{c}<\mathrm{2}{c}\:\mathrm{and}\:\mathrm{probably}\:{c}+{c}<{c}^{\mathrm{2}} \:\Rightarrow\:{c}>\mathrm{2} \\ $$$${c}^{\mathrm{2}} +{c}^{\mathrm{2}} +{b}=\mathrm{13} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{the}\:\mathrm{boy}\:\mathrm{now}\:\mathrm{wears}\:\mathrm{shoes},\:\mathrm{they}\:\mathrm{rather} \\ $$$$\mathrm{look}\:\mathrm{multiplicated}\:\mathrm{again},\:\mathrm{and}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{boy}'\mathrm{s} \\ $$$$\mathrm{feet}\:\mathrm{out}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shoes}\:\mathrm{we}\:\mathrm{will}\:\mathrm{have}\:\mathrm{to}\:\mathrm{extract} \\ $$$$\mathrm{the}\:\mathrm{root}\:\Rightarrow\:\sqrt[{{b}}]{{s}_{{r}} {s}_{{l}} }.\:\mathrm{a}\:\mathrm{boy}\:\mathrm{looking}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one} \\ $$$$\mathrm{wearing}\:\mathrm{pink}\:\mathrm{shoes}\:\mathrm{only}\:\mathrm{exists}\:\mathrm{in}\:\mathrm{somebody}'\mathrm{s} \\ $$$$\mathrm{weird}\:\mathrm{dreams}\:\Rightarrow\:\mathrm{i}\sqrt[{{b}}]{{s}_{{r}} {s}_{{l}} }\:\mathrm{and}\:\mathrm{he}\:\mathrm{holds}\:\mathrm{2}\:\mathrm{cones}, \\ $$$${s}_{{r}} +\mathrm{i}\sqrt[{{b}}]{{s}_{{r}} {s}_{{l}} }×{c}^{\mathrm{3}} =? \\ $$$$\mathrm{this}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{using}\:\mathrm{elementary} \\ $$$$\mathrm{calculus}\:\mathrm{but}\:\mathrm{using}\:\mathrm{the}\:\mathrm{secret}\:\mathrm{formula}\:\mathrm{of} \\ $$$$\mathrm{Ramanjuan\&Ostrogradski}\:\mathrm{at}\:\mathrm{least}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{the}\:\mathrm{Shoemaker}−\mathrm{Conestant}\:\xi_{\mathrm{0}} =\mathrm{43} \\ $$
Commented by behi83417@gmail.com last updated on 08/Nov/18
6(shoes)=30⇒shoe=5 2(man)+2(shoe)=20⇒man=5 4(paper)+man=13⇒paper=2 now: (shoe)+(man+2shoes+2paper)×paper= =5+(5+2×5+2×2)×2= 43.
$$\mathrm{6}\left({shoes}\right)=\mathrm{30}\Rightarrow{shoe}=\mathrm{5} \\ $$$$\mathrm{2}\left({man}\right)+\mathrm{2}\left({shoe}\right)=\mathrm{20}\Rightarrow{man}=\mathrm{5} \\ $$$$\mathrm{4}\left({paper}\right)+{man}=\mathrm{13}\Rightarrow{paper}=\mathrm{2} \\ $$$${now}: \\ $$$$\left({shoe}\right)+\left({man}+\mathrm{2}{shoes}+\mathrm{2}{paper}\right)×{paper}= \\ $$$$=\mathrm{5}+\left(\mathrm{5}+\mathrm{2}×\mathrm{5}+\mathrm{2}×\mathrm{2}\right)×\mathrm{2}=\:\mathrm{43}. \\ $$
Answered by ajfour last updated on 08/Nov/18
((30+20)/2)−13 .
$$\frac{\mathrm{30}+\mathrm{20}}{\mathrm{2}}−\mathrm{13}\:. \\ $$

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