Puzzling Asked on March 26, 2021
Papa Smurf gathered some berries from the forest to the village and smurfs ate all those berries:
How many smurfs are there?
Reference: Bilim ve Teknik Dergisi 2018-08
Let $N$ be the number of smurfs.
The information we are given implies that one smurf ate $1/5$ of the total berries, and that each of the other $N-1$ smurfs ate at least $1/11$ of the total berries. This means we must have $$ frac{1}{5}+(N-1)frac{1}{11} leq 1, $$ and so $Nleq 9$.
The the smurf who ate the second-most at strictly less than $frac{1}{5}$ of the total, and each of the $N-2$ smurfs (other than those who at the most and second most) ate at most $frac{1}{10}$ of the total. This means we must have $$ frac{1}{5}+frac{1}{5}+(N-2)frac{1}{10}>1, $$ and so $Ngeq 9$.
Taken together, these bounds imply there must be $9$ smurfs.
Correct answer by Julian Rosen on March 26, 2021
I have an answer that works, but no proof of uniqueness yet.
Answered by jamisans on March 26, 2021
"The smurf who ate the most actually ate one-fourth of the berries eaten by the rest of the smurfs." -> This sentence implies that the smurth who ate the most ate 1/6 of the total berries (so there are at least 6 smurths).
"The smurf who ate the third-most actually ate one-ninth of the berries eaten by the rest of the smurfs" -> So the smurth who ate the third-most actually ate one-tenth of the total berries. So someone ate between 1/6 and 1/10.
"The smurf who ate the least actually ate one-tenth of the berries eaten by the rest of the smurfs" -> So everyone ate less than 1/11.
Let x the quantity eaten by the second-most (1/6 > x > 1/10). Let y the quantity eaten by the N-4 others smurfs ( (N-4)/11 < y < (N-4)/10 )
So the total quantity is x + y + 1/6 + 1/10 + 1/11 And so, we search N so that
1/6 + (N-4)/10 + 1/6 + 1/10 + 1/11 > 1 > 1/10 + (N-4)/11 + 1/6 + 1/10 + 1/11
So, N=9. There are 9 smurfs.
Answered by user51199 on March 26, 2021
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