Pie problem or a pi problem?

at a party they offer 100 guests a pie and distribute it as follows: the first guest has 1%, the second has 2%, and so on. So the question is: which guest will get the most (%) pie of the remaining?

9 25 30 50 100 1 10

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2 solutions

Pop Wong
Mar 17, 2021

Let G n G_n be the partition of pie that n t h n^{th} person get, F n F_n be the remaining Pie after n t h n^{th} person get their pie.

F n = 100 n 100 F n 1 ( 1 ) G n = n 100 F n 1 ( 2 ) F_n = \cfrac{100-n}{100} \cdot F_{n-1} \hspace{20mm} (1) \\ G_n = \cfrac{n}{100} \cdot F_{n-1} \hspace{26mm} (2) Rearrange (1),

F n 1 = 100 100 n F n ( 3 ) F_{n-1} = \cfrac{100}{100-n} \cdot F_{n} \hspace{20mm} (3)

Compare G n + 1 G_{n+1} and G n G_{n} ,

H ( n ) = G n + 1 G n = n + 1 100 F n n 100 F n 1 = n + 1 100 F n n 100 n F n sub (3) = ( n + 1 ) ( 100 n ) 100 n = n 2 + 99 n + 100 100 n H is monotonic decreasing as ( H ( n ) = 1 n 2 1 100 < 0 ) \begin{aligned} H(n) &= \cfrac{G_{n+1}}{G_n} = \cfrac{ \cfrac{n+1}{100} \cdot F_{n} } { \cfrac{n}{100} \cdot F_{n-1} } \hspace{10mm} \\ &=\cfrac{ \cfrac{n+1}{100} \cdot \cancel{F_{n}} } { \cfrac{n}{100-n} \cdot \cancel{ F_{n} } } \hspace{10mm} \text{sub (3)} \\ &=\cfrac{ (n+1)(100-n) }{ 100n } \\ &=\cfrac{ -n^2+99n+100 }{ 100n } \hspace{10mm} \text{H is monotonic decreasing as } \left( H'(n) = -\cfrac{1}{n^2} - \cfrac{1}{100} < 0 \right) \end{aligned}

With H ( 1 ) = 198 100 = 1.98 H(1) = \cfrac{198}{100} = 1.98 and H H is monotonic decreasing, we check which n n will make H ( n ) < 1 H(n) < 1 ; before that G n < G n + 1 G_{n} < G_{n+1}

H ( n ) = n 2 + 99 n + 100 100 n < 1 n 2 + 99 n + 100 < 100 n n 2 + n 100 > 0 n = 1 + 1 + 400 2 > 1 + 400 2 = 1 + 20 2 = 9.5 neglect the -ve \begin{aligned} H(n) &=\cfrac{ -n^2+99n+100 }{ 100n } < 1 \\ &\implies -n^2+99n+100 < 100n \\ &\implies n^2+n-100 > 0 \\ &\implies n = \cfrac{-1+\sqrt{1+400}}{2} > \cfrac{-1+\sqrt{400}}{2} = \cfrac{-1+ 20}{2} = 9.5 \hspace{5mm} \text{neglect the -ve } \\ \end{aligned}

It means when n > 9.5 , H ( n ) = G n + 1 G n < 1 n>9.5, H(n) = \cfrac{G_{n+1}}{G_{n}}< 1 which means when n 10 n\geq 10 next person will get less pie than previous person.

Therefore, H ( 9 ) = G 10 G 9 > 1 G 10 > G 9 H(9) = \cfrac{G_{10}}{G_{9}} > 1 \rightarrow G_{10} > G_{9} \therefore the 1 0 t h \boxed{10^{th} } person will get the most pie.

Good Solution. Thanks for posting

Elijah Frank - 2 months, 3 weeks ago
Elijah Frank
Mar 17, 2021

the formula and logic of this problem is the following: We define all this as factorial since the more we increase the number of guests we increase the multiplication of fractions for each guest, that is, the first is 1% of the whole while the second is already going up to 99 100 \frac{99}{100} over 2% and so on. Therefore we have that the total minus the difference of the first guest = 99 !, then on this we have the 100 guests minus the number of the guest that we want to know about his percentage (k) all that less the percentage of 1 100 \frac{1}{100} or - 10 10 0 k \frac{10}{100^k} . We obtain 99 ! 100 k ! \frac{99!}{100-k!} - k 10 0 k \frac{k}{100^k} so we can do 2 things We can do two things which is to divide the index k + 1 to know the ratio in which it will fall in this case if it is greater than 1 or 1 that will be our guest with most of there the graph referring to the percentage with the guests falls . another way but more delayed is to try with the options in this case the result is 10 therefore: 99 ! 100 10 ! \frac{99!}{100-10!} - 10 10 0 1 0 \frac{10}{100^10} = 6.28%, then drops (just a fun fact 6.28 is π*2).

@Elijah Frank I think in the question you should add "of the remaining" near 2% so that it would be more clear.

Omek K - 2 months, 3 weeks ago

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Thanks I would change it

Elijah Frank - 2 months, 3 weeks ago

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