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Question Number 58663 by mr W last updated on 27/Apr/19

Answered by ajfour last updated on 27/Apr/19

$${100\mathrm{T}}_{\mathrm{n}+1}=(\mathrm{n}+1)({\mathrm{T}}_{\mathrm{n}}+{\mathrm{T}}_{\mathrm{n}-1}+...+{\mathrm{T}}_1)$$$$\mathrm{and}{\mathrm{T}}_1+{\mathrm{T}}_2+....+{\mathrm{T}}_{\mathrm{n}}+...+{\mathrm{T}}_{\mathrm{N}}=1$$$$....$$

Answered by MJS last updated on 27/Apr/19

$$\mathrm{guest}10\mathrm{gets}\frac{245373636545037}{39062500000000}\approx 6.28157\mathrm{\%}$$$$\mathrm{interestingly}\mathrm{this}\mathrm{is}1.99948\pi$$$$\mathrm{guest}25\mathrm{is}\mathrm{the}\mathrm{last}\mathrm{one}\mathrm{to}\mathrm{get}\mathrm{at}\mathrm{least}1\mathrm{\%}$$$$\mathrm{if}\mathrm{the}\mathrm{pie}\mathrm{is}\mathrm{a}\mathrm{big}\mathrm{one}\mathrm{for}100\mathrm{persons},\mathrm{say}$$$$\mathrm{it}\mathrm{weighs}25\mathrm{kg}\mathrm{then}\mathrm{guest}40\mathrm{is}\mathrm{the}\mathrm{last}\mathrm{one}$$$$\mathrm{to}\mathrm{get}\mathrm{at}\mathrm{least}1\mathrm{g}$$$$\mathrm{the}\mathrm{first}11\mathrm{guests}\mathrm{get}\mathrm{almost}50\mathrm{\%}$$$$\mathrm{the}\mathrm{guests}29-100\mathrm{have}\mathrm{to}\mathrm{share}\approx 1\mathrm{\%}$$

Commented by peter frank last updated on 30/Apr/19

$$howdidyougetthose$$$$largenumber???$$

Commented by MJS last updated on 01/May/19

$$\mathrm{by}\mathrm{using}+-\times /$$

Commented by peter frank last updated on 01/May/19

$$sorrysirexplainfirst$$$$line$$

Commented by MJS last updated on 01/May/19

$$\mathrm{pie}100\mathrm{\%};{\mathrm{guest}}_{1}1\mathrm{\%}$$$$\mathrm{pie}=100\mathrm{\%}-1\mathrm{\%}=99\mathrm{\%};{\mathrm{guest}}_{2}2\mathrm{\%}\mathrm{of}99\mathrm{\%}=\frac{99}{50}\mathrm{\%}=1.98\mathrm{\%}$$$$\mathrm{pie}=99\mathrm{\%}-\frac{99}{50}\mathrm{\%}=\frac{4851}{50}\mathrm{\%};{\mathrm{guest}}_{3}3\mathrm{\%}\mathrm{of}\frac{4851}{50}\mathrm{\%}=\frac{14553}{5000}\mathrm{\%}\approx 2.91\mathrm{\%}$$$$\mathrm{p}=\frac{470547}{5000}\mathrm{\%};{\mathrm{g}}_4=\frac{470547}{125000}\mathrm{\%}\approx 3.76\mathrm{\%}$$$$\mathrm{p}=\frac{1411641}{15625}\mathrm{\%};{\mathrm{g}}_5=\frac{1411641}{312500}\mathrm{\%}\approx 4.52\mathrm{\%}$$$$\mathrm{p}=\frac{26821179}{312500}\mathrm{\%};{\mathrm{g}}_6=\frac{80463537}{15625000}\mathrm{\%}\approx 5.15\mathrm{\%}$$$$\mathrm{p}=\frac{1260595413}{15625000}\mathrm{\%};{\mathrm{g}}_7=\frac{8824167891}{1562500000}\mathrm{\%}\approx 5.65\mathrm{\%}$$$$\mathrm{p}=\frac{117235373409}{1562500000}\mathrm{\%};{\mathrm{g}}_8=\frac{117235373409}{19531250000}\mathrm{\%}\approx 6.00\mathrm{\%}$$$$\mathrm{p}=\frac{2696413588407}{39062500000}\mathrm{\%};{\mathrm{g}}_9=\frac{24267722295663}{3906250000000}\mathrm{\%}\approx 6.21\mathrm{\%}$$$$\mathrm{p}=\frac{245373636545037}{3906250000000}\mathrm{\%};{\mathrm{g}}_{10}=\frac{245373636545037}{39062500000000}\mathrm{\%}\approx 6.28\mathrm{\%}$$$$\mathrm{p}=\frac{2208362728905333}{39062500000000}\mathrm{\%};{\mathrm{g}}_{11}=\frac{24291990017958663}{3906250000000000}\mathrm{\%}\approx 6.22\mathrm{\%}$$

Commented by mr W last updated on 06/May/19

$$thanksforthesolutionsir!ihavetried$$$$tousethesamemethodtogeta$$$$generalform.$$

Answered by mr W last updated on 07/May/19

$$guest1gets\frac{1}{100},$$$$remainingpie:(1-\frac{1}{100})\times 1=\frac{99}{100}$$$$$$$$guest2gets\frac{2}{100}\times \frac{99}{100},$$$$remainingpie:(1-\frac{2}{100})\times \frac{99}{100}=\frac{98}{100}\times \frac{99}{100}$$$$$$$$guest3gets\frac{3}{100}\times \frac{98}{100}\times \frac{99}{100},$$$$remainingpie:(1-\frac{3}{100})\times \frac{98}{100}\times \frac{99}{100}=\frac{97}{100}\times \frac{98}{100}\times \frac{99}{100}$$$$...$$$$guestngets\frac{n}{100}\times \frac{99-n+2}{100}\times ...\times \frac{97}{100}\times \frac{98}{100}\times \frac{99}{100}$$$$remainingpie:\frac{100-n}{100}\times \frac{99-n+2}{100}\times ...\times \frac{97}{100}\times \frac{98}{100}\times \frac{99}{100}$$$$...$$$$T_n=\frac{n}{100}\times \frac{99-n+2}{100}\times ...\times \frac{97}{100}\times \frac{98}{100}\times \frac{99}{100}$$$$T_n=\frac{n(99-n+2)...97\times 98\times 99}{{100}^n}$$$$T_n=\frac{n(99-n+1)!(99-n+2)...97\times 98\times 99\times 100}{{100}^{n+1}(99-n+1)!}$$$$\Rightarrow T_n=\frac{100!n}{{100}^{n+1}(100-n)!}$$$$$$$$T_{n}ismaximumwhen{T}_{n-1}<T_n>T_{n+1},i.e.$$$$\frac{100!n}{{100}^{n+1}(100-n)!}>\frac{100!(n+1)}{{100}^{n+2}(100-n-1)!}$$$$\frac{n}{100-n}>\frac{n+1}{100}$$$$100n>(n+1)(100-n)$$$$100n>100n+100-n^2-n$$$$0>100-n^2-n$$$$n^2+n-100>0$$$$\Rightarrow n>\frac{-1+\sqrt{1+400}}{2}\approx 9.5$$$$i.e.the10thguestgetsthelargestpiece.$$$$T_{10}=\frac{100!10}{{100}^{11}90!}=\frac{99\times 97\times ...\times 92\times 91}{{10}^{19}}$$$$=\frac{62815650955529472}{10000000000000000000}$$$$\approx 6.28\mathrm{\%}$$

Commented by MJS last updated on 06/May/19

$$\mathrm{great}!$$

Commented by peter frank last updated on 19/May/19

$$thankyou$$

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