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Question-58663

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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
100T_(n+1) =(n+1)(T_n +T_(n−1) +...+T_1 ) and T_1 +T_2 +....+T_n +...+T_N = 1 ....
$$\:\mathrm{100T}_{\mathrm{n}+\mathrm{1}} =\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{T}_{\mathrm{n}} +\mathrm{T}_{\mathrm{n}−\mathrm{1}} +…+\mathrm{T}_{\mathrm{1}} \right) \\ $$$$\:\mathrm{and}\:\:\mathrm{T}_{\mathrm{1}} +\mathrm{T}_{\mathrm{2}} +….+\mathrm{T}_{\mathrm{n}} +…+\mathrm{T}_{\mathrm{N}} =\:\mathrm{1} \\ $$$$…. \\ $$
Answered by MJS last updated on 27/Apr/19
guest 10 gets ((245 373 636 545 037)/(39 062 500 000 000))≈6.28157% interestingly this is 1.99948π guest 25 is the last one to get at least 1% if the pie is a big one for 100 persons, say it weighs 25kg then guest 40 is the last one to get at least 1g the first 11 guests get almost 50% the guests 29−100 have to share ≈1%
$$\mathrm{guest}\:\mathrm{10}\:\mathrm{gets}\:\frac{\mathrm{245}\:\mathrm{373}\:\mathrm{636}\:\mathrm{545}\:\mathrm{037}}{\mathrm{39}\:\mathrm{062}\:\mathrm{500}\:\mathrm{000}\:\mathrm{000}}\approx\mathrm{6}.\mathrm{28157\%} \\ $$$$\mathrm{interestingly}\:\mathrm{this}\:\mathrm{is}\:\mathrm{1}.\mathrm{99948}\pi \\ $$$$\mathrm{guest}\:\mathrm{25}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{one}\:\mathrm{to}\:\mathrm{get}\:\mathrm{at}\:\mathrm{least}\:\mathrm{1\%} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{pie}\:\mathrm{is}\:\mathrm{a}\:\mathrm{big}\:\mathrm{one}\:\mathrm{for}\:\mathrm{100}\:\mathrm{persons},\:\mathrm{say} \\ $$$$\mathrm{it}\:\mathrm{weighs}\:\mathrm{25kg}\:\mathrm{then}\:\mathrm{guest}\:\mathrm{40}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{one} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{at}\:\mathrm{least}\:\mathrm{1g} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{11}\:\mathrm{guests}\:\mathrm{get}\:\mathrm{almost}\:\mathrm{50\%} \\ $$$$\mathrm{the}\:\mathrm{guests}\:\mathrm{29}−\mathrm{100}\:\mathrm{have}\:\mathrm{to}\:\mathrm{share}\:\approx\mathrm{1\%} \\ $$
Commented by peter frank last updated on 30/Apr/19
how did you get those large number???
$${how}\:{did}\:{you}\:{get}\:{those} \\ $$$${large}\:{number}??? \\ $$
Commented by MJS last updated on 01/May/19
by using +−×/
$$\mathrm{by}\:\mathrm{using}\:+−×/ \\ $$
Commented by peter frank last updated on 01/May/19
sorry sir explain first line
$${sorry}\:{sir}\:{explain}\:{first}\: \\ $$$${line} \\ $$
Commented by MJS last updated on 01/May/19
pie 100%; guest_1 1% pie=100%−1%= 99%; guest_2 2% of 99%=((99)/(50))%=1.98% pie=99%−((99)/(50))%=((4 851)/(50))%; guest_3 3% of ((4 851)/(50))%=((14 553)/(5 000))%≈2.91% p=((470 547)/(5 000))%; g_4 =((470 547)/(125 000))%≈3.76% p=((1 411 641)/(15 625))%; g_5 =((1 411 641)/(312 500))%≈4.52% p=((26 821 179)/(312 500))%; g_6 =((80 463 537)/(15 625 000))%≈5.15% p=((1 260 595 413)/(15 625 000))%; g_7 =((8 824 167 891)/(1 562 500 000))%≈5.65% p=((117 235 373 409)/(1 562 500 000))%; g_8 =((117 235 373 409)/(19 531 250 000))%≈6.00% p=((2 696 413 588 407)/(39 062 500 000))%; g_9 =((24 267 722 295 663)/(3 906 250 000 000))%≈6.21% p=((245 373 636 545 037)/(3 906 250 000 000))%; g_(10) =((245 373 636 545 037)/(39 062 500 000 000))%≈6.28% p=((2 208 362 728 905 333)/(39 062 500 000 000))%; g_(11) =((24 291 990 017 958 663)/(3 906 250 000 000 000))%≈6.22%
$$\mathrm{pie}\:\mathrm{100\%};\:\mathrm{guest}_{\mathrm{1}} \:\mathrm{1\%} \\ $$$$\mathrm{pie}=\mathrm{100\%}−\mathrm{1\%}=\:\mathrm{99\%};\:\mathrm{guest}_{\mathrm{2}} \:\mathrm{2\%}\:\mathrm{of}\:\mathrm{99\%}=\frac{\mathrm{99}}{\mathrm{50}}\%=\mathrm{1}.\mathrm{98\%} \\ $$$$\mathrm{pie}=\mathrm{99\%}−\frac{\mathrm{99}}{\mathrm{50}}\%=\frac{\mathrm{4}\:\mathrm{851}}{\mathrm{50}}\%;\:\mathrm{guest}_{\mathrm{3}} \:\mathrm{3\%}\:\mathrm{of}\:\frac{\mathrm{4}\:\mathrm{851}}{\mathrm{50}}\%=\frac{\mathrm{14}\:\mathrm{553}}{\mathrm{5}\:\mathrm{000}}\%\approx\mathrm{2}.\mathrm{91\%} \\ $$$$\mathrm{p}=\frac{\mathrm{470}\:\mathrm{547}}{\mathrm{5}\:\mathrm{000}}\%;\:\mathrm{g}_{\mathrm{4}} =\frac{\mathrm{470}\:\mathrm{547}}{\mathrm{125}\:\mathrm{000}}\%\approx\mathrm{3}.\mathrm{76\%} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\:\mathrm{411}\:\mathrm{641}}{\mathrm{15}\:\mathrm{625}}\%;\:\mathrm{g}_{\mathrm{5}} =\frac{\mathrm{1}\:\mathrm{411}\:\mathrm{641}}{\mathrm{312}\:\mathrm{500}}\%\approx\mathrm{4}.\mathrm{52\%} \\ $$$$\mathrm{p}=\frac{\mathrm{26}\:\mathrm{821}\:\mathrm{179}}{\mathrm{312}\:\mathrm{500}}\%;\:\mathrm{g}_{\mathrm{6}} =\frac{\mathrm{80}\:\mathrm{463}\:\mathrm{537}}{\mathrm{15}\:\mathrm{625}\:\mathrm{000}}\%\approx\mathrm{5}.\mathrm{15\%} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\:\mathrm{260}\:\mathrm{595}\:\mathrm{413}}{\mathrm{15}\:\mathrm{625}\:\mathrm{000}}\%;\:\mathrm{g}_{\mathrm{7}} =\frac{\mathrm{8}\:\mathrm{824}\:\mathrm{167}\:\mathrm{891}}{\mathrm{1}\:\mathrm{562}\:\mathrm{500}\:\mathrm{000}}\%\approx\mathrm{5}.\mathrm{65\%} \\ $$$$\mathrm{p}=\frac{\mathrm{117}\:\mathrm{235}\:\mathrm{373}\:\mathrm{409}}{\mathrm{1}\:\mathrm{562}\:\mathrm{500}\:\mathrm{000}}\%;\:\mathrm{g}_{\mathrm{8}} =\frac{\mathrm{117}\:\mathrm{235}\:\mathrm{373}\:\mathrm{409}}{\mathrm{19}\:\mathrm{531}\:\mathrm{250}\:\mathrm{000}}\%\approx\mathrm{6}.\mathrm{00\%} \\ $$$$\mathrm{p}=\frac{\mathrm{2}\:\mathrm{696}\:\mathrm{413}\:\mathrm{588}\:\mathrm{407}}{\mathrm{39}\:\mathrm{062}\:\mathrm{500}\:\mathrm{000}}\%;\:\mathrm{g}_{\mathrm{9}} =\frac{\mathrm{24}\:\mathrm{267}\:\mathrm{722}\:\mathrm{295}\:\mathrm{663}}{\mathrm{3}\:\mathrm{906}\:\mathrm{250}\:\mathrm{000}\:\mathrm{000}}\%\approx\mathrm{6}.\mathrm{21\%} \\ $$$$\mathrm{p}=\frac{\mathrm{245}\:\mathrm{373}\:\mathrm{636}\:\mathrm{545}\:\mathrm{037}}{\mathrm{3}\:\mathrm{906}\:\mathrm{250}\:\mathrm{000}\:\mathrm{000}}\%;\:\mathrm{g}_{\mathrm{10}} =\frac{\mathrm{245}\:\mathrm{373}\:\mathrm{636}\:\mathrm{545}\:\mathrm{037}}{\mathrm{39}\:\mathrm{062}\:\mathrm{500}\:\mathrm{000}\:\mathrm{000}}\%\approx\mathrm{6}.\mathrm{28\%} \\ $$$$\mathrm{p}=\frac{\mathrm{2}\:\mathrm{208}\:\mathrm{362}\:\mathrm{728}\:\mathrm{905}\:\mathrm{333}}{\mathrm{39}\:\mathrm{062}\:\mathrm{500}\:\mathrm{000}\:\mathrm{000}}\%;\:\mathrm{g}_{\mathrm{11}} =\frac{\mathrm{24}\:\mathrm{291}\:\mathrm{990}\:\mathrm{017}\:\mathrm{958}\:\mathrm{663}}{\mathrm{3}\:\mathrm{906}\:\mathrm{250}\:\mathrm{000}\:\mathrm{000}\:\mathrm{000}}\%\approx\mathrm{6}.\mathrm{22\%} \\ $$
Commented by mr W last updated on 06/May/19
thanks for the solution sir! i have tried to use the same method to get a general form.
$${thanks}\:{for}\:{the}\:{solution}\:{sir}!\:{i}\:{have}\:{tried} \\ $$$${to}\:{use}\:{the}\:{same}\:{method}\:{to}\:{get}\:{a}\: \\ $$$${general}\:{form}. \\ $$
Answered by mr W last updated on 07/May/19
guest 1 gets (1/(100)), remaining pie: (1−(1/(100)))×1=((99)/(100)) guest 2 gets (2/(100))×((99)/(100)), remaining pie: (1−(2/(100)))×((99)/(100))=((98)/(100))×((99)/(100)) guest 3 gets (3/(100))×((98)/(100))×((99)/(100)), remaining pie: (1−(3/(100)))×((98)/(100))×((99)/(100))=((97)/(100))×((98)/(100))×((99)/(100)) ... guest n gets (n/(100))×((99−n+2)/(100))×...×((97)/(100))×((98)/(100))×((99)/(100)) remaining pie: ((100−n)/(100))×((99−n+2)/(100))×...×((97)/(100))×((98)/(100))×((99)/(100)) ... T_n =(n/(100))×((99−n+2)/(100))×...×((97)/(100))×((98)/(100))×((99)/(100)) T_n =((n(99−n+2)...97×98×99)/(100^n )) T_n =((n(99−n+1)!(99−n+2)...97×98×99×100)/(100^(n+1) (99−n+1)!)) ⇒T_n =((100!n)/(100^(n+1) (100−n)!)) T_n is maximum when T_(n−1) <T_n >T_(n+1) , i.e. ((100!n)/(100^(n+1) (100−n)!))>((100!(n+1))/(100^(n+2) (100−n−1)!)) (n/(100−n))>((n+1)/(100)) 100n>(n+1)(100−n) 100n>100n+100−n^2 −n 0>100−n^2 −n n^2 +n−100>0 ⇒n>((−1+(√(1+400)))/2)≈9.5 i.e. the 10th guest gets the largest piece. T_(10) =((100!10)/(100^(11) 90!))=((99×97×...×92×91)/(10^(19) )) =((62815650955529472)/(10000000000000000000)) ≈6.28%
$${guest}\:\mathrm{1}\:{gets}\:\frac{\mathrm{1}}{\mathrm{100}}, \\ $$$${remaining}\:{pie}:\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}}\right)×\mathrm{1}=\frac{\mathrm{99}}{\mathrm{100}} \\ $$$$ \\ $$$${guest}\:\mathrm{2}\:{gets}\:\frac{\mathrm{2}}{\mathrm{100}}×\frac{\mathrm{99}}{\mathrm{100}}, \\ $$$${remaining}\:{pie}:\:\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{100}}\right)×\frac{\mathrm{99}}{\mathrm{100}}=\frac{\mathrm{98}}{\mathrm{100}}×\frac{\mathrm{99}}{\mathrm{100}} \\ $$$$ \\ $$$${guest}\:\mathrm{3}\:{gets}\:\frac{\mathrm{3}}{\mathrm{100}}×\frac{\mathrm{98}}{\mathrm{100}}×\frac{\mathrm{99}}{\mathrm{100}}, \\ $$$${remaining}\:{pie}:\:\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{100}}\right)×\frac{\mathrm{98}}{\mathrm{100}}×\frac{\mathrm{99}}{\mathrm{100}}=\frac{\mathrm{97}}{\mathrm{100}}×\frac{\mathrm{98}}{\mathrm{100}}×\frac{\mathrm{99}}{\mathrm{100}} \\ $$$$… \\ $$$${guest}\:{n}\:{gets}\:\frac{{n}}{\mathrm{100}}×\frac{\mathrm{99}−{n}+\mathrm{2}}{\mathrm{100}}×…×\frac{\mathrm{97}}{\mathrm{100}}×\frac{\mathrm{98}}{\mathrm{100}}×\frac{\mathrm{99}}{\mathrm{100}} \\ $$$${remaining}\:{pie}:\:\:\frac{\mathrm{100}−{n}}{\mathrm{100}}×\frac{\mathrm{99}−{n}+\mathrm{2}}{\mathrm{100}}×…×\frac{\mathrm{97}}{\mathrm{100}}×\frac{\mathrm{98}}{\mathrm{100}}×\frac{\mathrm{99}}{\mathrm{100}} \\ $$$$… \\ $$$${T}_{{n}} =\frac{{n}}{\mathrm{100}}×\frac{\mathrm{99}−{n}+\mathrm{2}}{\mathrm{100}}×…×\frac{\mathrm{97}}{\mathrm{100}}×\frac{\mathrm{98}}{\mathrm{100}}×\frac{\mathrm{99}}{\mathrm{100}} \\ $$$${T}_{{n}} =\frac{{n}\left(\mathrm{99}−{n}+\mathrm{2}\right)…\mathrm{97}×\mathrm{98}×\mathrm{99}}{\mathrm{100}^{{n}} } \\ $$$${T}_{{n}} =\frac{{n}\left(\mathrm{99}−{n}+\mathrm{1}\right)!\left(\mathrm{99}−{n}+\mathrm{2}\right)…\mathrm{97}×\mathrm{98}×\mathrm{99}×\mathrm{100}}{\mathrm{100}^{{n}+\mathrm{1}} \left(\mathrm{99}−{n}+\mathrm{1}\right)!} \\ $$$$\Rightarrow{T}_{{n}} =\frac{\mathrm{100}!{n}}{\mathrm{100}^{{n}+\mathrm{1}} \left(\mathrm{100}−{n}\right)!} \\ $$$$ \\ $$$${T}_{{n}} \:{is}\:{maximum}\:{when}\:{T}_{{n}−\mathrm{1}} <{T}_{{n}} >{T}_{{n}+\mathrm{1}} ,\:{i}.{e}. \\ $$$$\frac{\mathrm{100}!{n}}{\mathrm{100}^{{n}+\mathrm{1}} \left(\mathrm{100}−{n}\right)!}>\frac{\mathrm{100}!\left({n}+\mathrm{1}\right)}{\mathrm{100}^{{n}+\mathrm{2}} \left(\mathrm{100}−{n}−\mathrm{1}\right)!} \\ $$$$\frac{{n}}{\mathrm{100}−{n}}>\frac{{n}+\mathrm{1}}{\mathrm{100}} \\ $$$$\mathrm{100}{n}>\left({n}+\mathrm{1}\right)\left(\mathrm{100}−{n}\right) \\ $$$$\mathrm{100}{n}>\mathrm{100}{n}+\mathrm{100}−{n}^{\mathrm{2}} −{n} \\ $$$$\mathrm{0}>\mathrm{100}−{n}^{\mathrm{2}} −{n} \\ $$$${n}^{\mathrm{2}} +{n}−\mathrm{100}>\mathrm{0} \\ $$$$\Rightarrow{n}>\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{400}}}{\mathrm{2}}\approx\mathrm{9}.\mathrm{5} \\ $$$${i}.{e}.\:{the}\:\mathrm{10}{th}\:{guest}\:{gets}\:{the}\:{largest}\:{piece}. \\ $$$${T}_{\mathrm{10}} =\frac{\mathrm{100}!\mathrm{10}}{\mathrm{100}^{\mathrm{11}} \mathrm{90}!}=\frac{\mathrm{99}×\mathrm{97}×…×\mathrm{92}×\mathrm{91}}{\mathrm{10}^{\mathrm{19}} } \\ $$$$=\frac{\mathrm{62815650955529472}}{\mathrm{10000000000000000000}} \\ $$$$\approx\mathrm{6}.\mathrm{28\%} \\ $$
Commented by MJS last updated on 06/May/19
great!
$$\mathrm{great}! \\ $$
Commented by peter frank last updated on 19/May/19
thank you
$${thank}\:{you} \\ $$

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