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?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Good Solution. Thanks for posting
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 1 0 0 9 9 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 0 0 1 or - 1 0 0 k 1 0 . We obtain 1 0 0 − k ! 9 9 ! - 1 0 0 k 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: 1 0 0 − 1 0 ! 9 9 ! - 1 0 0 1 0 1 0 = 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.
Problem Loading...
Note Loading...
Set Loading...
Let G n be the partition of pie that n t h person get, F n be the remaining Pie after n t h person get their pie.
F n = 1 0 0 1 0 0 − n ⋅ F n − 1 ( 1 ) G n = 1 0 0 n ⋅ F n − 1 ( 2 ) Rearrange (1),
F n − 1 = 1 0 0 − n 1 0 0 ⋅ F n ( 3 )
Compare G n + 1 and G n ,
H ( n ) = G n G n + 1 = 1 0 0 n ⋅ F n − 1 1 0 0 n + 1 ⋅ F n = 1 0 0 − n n ⋅ F n 1 0 0 n + 1 ⋅ F n sub (3) = 1 0 0 n ( n + 1 ) ( 1 0 0 − n ) = 1 0 0 n − n 2 + 9 9 n + 1 0 0 H is monotonic decreasing as ( H ′ ( n ) = − n 2 1 − 1 0 0 1 < 0 )
With H ( 1 ) = 1 0 0 1 9 8 = 1 . 9 8 and H is monotonic decreasing, we check which n will make H ( n ) < 1 ; before that G n < G n + 1
H ( n ) = 1 0 0 n − n 2 + 9 9 n + 1 0 0 < 1 ⟹ − n 2 + 9 9 n + 1 0 0 < 1 0 0 n ⟹ n 2 + n − 1 0 0 > 0 ⟹ n = 2 − 1 + 1 + 4 0 0 > 2 − 1 + 4 0 0 = 2 − 1 + 2 0 = 9 . 5 neglect the -ve
It means when n > 9 . 5 , H ( n ) = G n G n + 1 < 1 which means when n ≥ 1 0 next person will get less pie than previous person.
Therefore, H ( 9 ) = G 9 G 1 0 > 1 → G 1 0 > G 9 ∴ the 1 0 t h person will get the most pie.