A pie
Hasmik invited several guests, and she knows that there will either be 7 or 8 people at the party. She wants to slice up a big pie into smaller pieces, not necessarily of the same size, such that regardless of how many people show up, she can serve the entire pie evenly to everyone.
What is the minimum number of pieces she will need to slice the big pie into?
The answer is 14.
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9 solutions
As always, to prove something is the minimum, we have to show that
1. It can be achieved - this is done.
2. No smaller number can be achieved - this has not been done.
It is easy to make this into a complete solution.
No piece can be bigger than 1/8th of the pie (if a piece was bigger than that, there'd be no way to accommodate 8 guests). Therefore, if 7 guests arrive then they must each be given at least two pieces. Combined with what you already demonstrated, that proves that 14 is minimal.
Small bit of pedantry: it's either 6 or 7 guests. "Everybody" includes the host.
But wait- there are 8 or 7 guest and SHE. With here there would be 9 or 8 people eating cake. So shouldn't she cut 9 slices?
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No I don't think so- there are 8 or 7 people, not 8 or 7 guests
I just tried to use my logical thinking. How is 14 the MINIMUM? I would rate this problem a 9.5 because it was kind of hard,
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i don't get the point here either,, she could just cut it evenly in 4 diameters with equal angle between them ?
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How would she divide the pie evenly between 7 people if she did this?
Can't we make 56 slices and give each one 8 or 7 slices depending if number of guest is 7 or 8?
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We can divide the pie as you say in 56 pieces, but the question ask for the minimum number of pieces needed. This division can be done using different sizes of slices. So if you cleverly cut the cake as shown on the solutions you don't have to make 56 pieces, with 14 pieces is enough and this number of slices is the smallest you can use.
Relevant wiki: Fair Division
Cut the pie into 8 equal sized pieces. Then take one of the 8 pieces and cut it into 7 equal sized pieces.
This leaves 7 pieces that are each 1/8 of the pie + 7 smaller pieces that together make up 1/8 of the pie. 7+7= a total of 14 pieces of pie.
In the event that 7 people come to the party each quest will get one of the larger pieces + one of the smaller pieces of pie.
In the event that 8 people come to the party 7 quests will each get 1 large piece of pie and one quest will get all 7 of the smaller pieces of pie.
As always, to prove something is the minimum, we have to show that
1. It can be achieved - this is done.
2. No smaller number can be achieved - this has not been done.
You are assuming that there must be some kind of structure to the way the pie is cut up, namely "take one of the 8 pieces and cut it into 7 equal sized pieces.". There is apriori no reason why each of the 8 pieces has to be a whole piece. In fact, you violate this assumption by later cutting one of them up into 7 pieces. Why can't it be possible that there is some other weird arrangement that satisfies the conditions?
This is a simply achieved solution to the problem. It seams apparent to me that it is not possible to divide the pie in less then 14 pieces and achieve the goal. Using a bit of reasoning. 1 piece of pie for each person is the smallest possible way to distribute any pie between any number of people. . By definition if each person gets a piece of pie there must by at least as many pieces of pie as there are people. Therefore, to divide the pie between 8 people would require at least 8 pieces. To divided the pie between 7 people would require at least 7 pieces. It is also apparent that if the portions of pie are to be equal for each person in a group of 8 or alternatively must equal for each person in the group of 7 then it is not possible for the portions to be equal for any individual person in both circumstances. i.e. the amount of pie that each and any individual person will receive must be a different portion when distributing by 8 compared when distributing by 7.
There will be ether 7 or 8 people are attending the party. The only possible way to have two different portions (1/7 &1/8) available to any individual is to have his or her portion divided into at least two pieces in at least one of the circumstances i.e. one piece of pie cannot be both 1/7 and 1/8 of the pie. There are at least seven individuals at the party,that would be the lowest possible number of individuals that require at least 2 portions of pie in order to have both 1/8 and 1/7 available to them. Seven people each have at least two pieces in one of the circumstances means the lowest possible number will be 7X2=14 pieces. It is not known if it is possible to actually achieve that number. It is clear that 14 is the lowest number that may or may not be possible. There cannot be any weird arrangement that make two options of seven equal sized pieces available to seven individuals in less then 14 pieces. If any individual has a piece of the pie that cannot be altered and we must change the size of their portion the only way to do this is give them each another piece of the pie now they have at least two pieces of pie..
To put this another way. When the pie is divided equally to 8 people. The lower possible number of pieces of pie is clearly 8 no distribution can reduce that number . The only possible way to give any of these 8 a larger portion of the pie is to cut up one or more of the 8 pieces and distribute it into the renaming people. If all the remaining people are to have the same portion after redistribution every one of the remaining people must receive at least one more piece of pie. If our goal is to have 7 people remaining after redistribution the lowest possible total pieces of pie would be 14 since each of the remaining 7 people would now have a least 2 pieces of pie.
It is very clear to me that this cannot be done in less then 14 pieces. Once a simple solution is determined that takes only 14 pieces that is all that is required.
First, let us assume that the pie needs to be wholly apportioned, i.e. without leftovers; else, the problem becomes trivial, as the hostess, could, for instance, slice the pie into any number of pieces but serve no pie to each guest, thus ensuring equal distribution.
Let the pie be divided into 7 equal-sized slices. Clearly, this is the minimum number of pieces needed to satisfy the case in which 7 people arrive at the party. In the case that 8 people arrive, however, one-eighth of the original pie will need to be constructed from portions of the 7 pieces. As each guest must receive an equal portion, an equal portion must be removed from each of the original 7 pieces to assemble an eighth portion—and such that each of the 8 portions is now one-eighth of the big pie. Hence, no fewer than 7 more slices must be made, for a total of 14 slices, minimum.
Upvoted for the genial idea of keeping all the pie for yourself, Sir.
Look. If you throw a party you want from the cake too. I was counting with her also involved. So that's unfair. In the text it says following: "she can serve the entire pie evenly to everyone." EVERYONE being the keyword.
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There are 7 or 8 people at the party, that includes the hostess.
I also upvoted since the problem is worded poorly and serving no pie is a solution. However, since the question says 'everyone' I made it 9 slices with the 9th divided since as Hasmik is present and everyone gets a slice, the counts would become 8-9 instead of 7-8.
7 parts of 1/8 of pie, and 7 parts of 1/56. 7 guests get 1/8+1/56 each, 8 guests get 1/8, one of them all small pieces.
The way the problem is written, the minimum is 8 pieces -- If 7 people show up, they each get 1/8th of the pie with none left. If 8 people show up, each gets 1/8th. The problem doesn't say that she intends to have no pie leftover.
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Thanks. Those who answered 8 have been marked correct.
I have edited the problem for clarity.
In future, if you report a problem directly, that will notify the problem creator and Brilliant staff.
How can we show that this is indeed the minimum? You have shown that it can be achieved, but still need to show that no smaller value works.
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If 13 pieces, one of 7 guest get 1 piece which is larger than 1/8. Then who get it when 8 guest come? It is impossible, so 14 is the minimum.
I must admit that this problem was created by Sharygin from Moscow who died in 2004.
Let the size of the cake be 56 units for convenience.
First, consider 7 people to serve. Since we need pieces that can be grouped into portions each of size 8, as a starting point we cut the pie into 7 such pieces.
To cater for 8, we need no piece to be larger than 7. So we need to cut each of the 7 pieces into at least two, giving a minimum 14 pieces. And 14 is indeed possible in a number of ways:
- 7 × ( 7 + 1 ) as given by others
- 7 + 1 + 6 + 2 + 5 + 3 + 4 + 4 + 3 + 5 + 2 + 6 + 1 + 7 (this way, each person's portion will be either a single piece or a pair of adjacent pieces).
I think it should be clarified in the question that the whole pie needs to be served. Otherwise 8 pieces is a valid solution, as you can just leave a piece uneaten if there are only 7 guests.
Thanks. Those who answered 8 have been marked correct.
I have edited the problem for clarity.
In future, if you report a problem directly, that will notify the problem creator and Brilliant staff.
Of course, in the real world, 8 is the correct answer anyway. If 7 guests come, the hostess joins in and has a piece along with the guests. If 8 come, she defers and says she's trying to cut back. In both cases, she serves the pie evenly and no pie is left over.
Why didn't slice the pie horizontally into 8 pieces and then distribute 8th piece into 7 if seven people showed up.
We can visualize 1/7 as 1/56+1/8 so each 1/7 can be cut into two pieces, one 1/56 and the other 1/8. If there are eight guests, seven of them get a slice of 1/8 each while the last guest gets 7 slices of 1/56 which totals 1/8. If there are seven guests, each gets a slice of 1/56 and a slice of 1/8. With 1/7 cut into two pieces, we have 7 × 2 = 1 4 pieces.
I started out by simply multiplying the two possible amounts of guests, so 7 x 8 = 56.
Only afterwards have I noticed that it needs to be the "minimum number of pieces".
Okay...
So I started halving: 3.5 x 4 = 14 <-- could work (more than 8).
1.75 x 2 = 3.5 <-- won't work, not enough (less than 7).
So, 14.
Let's prove that the minimum solution is actually the minimum.
In the minimum solution, there must be a way to group all the slices into equal portions of 1/7 of the whole pie (to satisfy the case of 7 people). The only question, then, is how many slices each 1/7-portion must contain to satisfy the 8-person case.
We can deduce that each 1/7-portion must contain at least 2 slices (for a total of 1 4 slices). Otherwise, at least one slice will be 1/7 of the pie, which is too big to fairly serve in the 8-person case.
Now we just have to show that a solution with 14 slices exists: 7 1/8-slices, plus the last 1/8-slice split into 7 equal mini-slices.
Relevant wiki: Fair Division
Should there be 8 people, 7 can take a large slice, and the last could take the 7 small slices.
Should there be 7 people, they could each take 1 small slice and 1 large slice.