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Question-58663




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  ....
100Tn+1=(n+1)(Tn+Tn1++T1)andT1+T2+.+Tn++TN=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%
guest10gets245373636545037390625000000006.28157%interestinglythisis1.99948πguest25isthelastonetogetatleast1%ifthepieisabigonefor100persons,sayitweighs25kgthenguest40isthelastonetogetatleast1gthefirst11guestsgetalmost50%theguests29100havetoshare1%
Commented by peter frank last updated on 30/Apr/19
how did you get those  large number???
howdidyougetthoselargenumber???
Commented by MJS last updated on 01/May/19
by using +−×/
byusing+×/
Commented by peter frank last updated on 01/May/19
sorry sir explain first   line
sorrysirexplainfirstline
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%
pie100%;guest11%pie=100%1%=99%;guest22%of99%=9950%=1.98%pie=99%9950%=485150%;guest33%of485150%=145535000%2.91%p=4705475000%;g4=470547125000%3.76%p=141164115625%;g5=1411641312500%4.52%p=26821179312500%;g6=8046353715625000%5.15%p=126059541315625000%;g7=88241678911562500000%5.65%p=1172353734091562500000%;g8=11723537340919531250000%6.00%p=269641358840739062500000%;g9=242677222956633906250000000%6.21%p=2453736365450373906250000000%;g10=24537363654503739062500000000%6.28%p=220836272890533339062500000000%;g11=242919900179586633906250000000000%6.22%
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.
thanksforthesolutionsir!ihavetriedtousethesamemethodtogetageneralform.
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%
guest1gets1100,remainingpie:(11100)×1=99100guest2gets2100×99100,remainingpie:(12100)×99100=98100×99100guest3gets3100×98100×99100,remainingpie:(13100)×98100×99100=97100×98100×99100guestngetsn100×99n+2100××97100×98100×99100remainingpie:100n100×99n+2100××97100×98100×99100Tn=n100×99n+2100××97100×98100×99100Tn=n(99n+2)97×98×99100nTn=n(99n+1)!(99n+2)97×98×99×100100n+1(99n+1)!Tn=100!n100n+1(100n)!TnismaximumwhenTn1<Tn>Tn+1,i.e.100!n100n+1(100n)!>100!(n+1)100n+2(100n1)!n100n>n+1100100n>(n+1)(100n)100n>100n+100n2n0>100n2nn2+n100>0n>1+1+40029.5i.e.the10thguestgetsthelargestpiece.T10=100!101001190!=99×97××92×911019=62815650955529472100000000000000000006.28%
Commented by MJS last updated on 06/May/19
great!
great!
Commented by peter frank last updated on 19/May/19
thank you
thankyou

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