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Some fishermen caught some fish. No one caught more than 20 fish. $a_i$ is the number of fishermen who caught at least $i$ fish. How many fish were caught?

So my guess is that the number of fish caught has to be $$ a_1+a_2+\ldots + a_{20}$$

I reason that if a fisher man caught $n$ fish then he will appear in the tally $n$ times.

Is this logic correct?

Castor Godinho
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  • Nice question! My first thought is that $a_i$ must be a descending sequence, because if a fisher catches at least $i+1$ fish, it's also true that they have caught at least $i$ fish. Therefore, if a fisher belongs to group $i+1$, then he also belongs to group $i$. – Matti P. Jun 19 '17 at 07:31
  • @MattiP Yes. I totally agree. – Castor Godinho Jun 19 '17 at 07:32

2 Answers2

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Yes, you are right. Here is a different (more concrete, I guess) way of seeing it:

Put out barrels numbered 1 through 20 and tell each fisherman to put their first fish in barrel 1, their second in barrel 2, and so on. Each fish gets put in a barrel, and the number of fish is therefore necessarily equal to the sum of fish in each barrel. However, we also have that $a_i$ is the number of fish in barrel $i$, since any fisherman who got at least $i$ fish will put one fish into barrel $i$. Thus we get your result.

Arthur
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You are right. I used a table to make the problem more clear to me.

Let $x_i$ be the number of fishermen who caught i fishes. Suppose we have the following table:

$$\begin{array}{|c|c|c|c|c|} \hline i&1&2&3&4&\ldots&20 \\ \hline x_i&x_1&x_2&x_3&x_4&\ldots&x_{20} \\ \hline a_i&a_1&a_2&a_3&a_4&\ldots&a_{20} \\ \hline\end{array}$$

The number of fishes which are caught by fishermen who caught 1 fish is $1\cdot x_1=(a_1-a_2)\cdot 1$

The number of fishes which are caught by fishermen who caught 2 fishes is $2\cdot x_2=(a_2-a_3)\cdot 2$

The number of fishes which are caught by fishermen who caught 3 fishes is $3\cdot x_3=(a_3-a_4)\cdot 3$

$\ldots$

The number of fishes which are caught by fishermen who caught 20 fishes is $20\cdot x_{20}=a_{20}\cdot 20$

Summing up the terms

$$(a_1-a_2)\cdot 1+(a_2-a_3)\cdot 2+(a_3-a_4)\cdot 3+\ldots -a_{20}\cdot 19+a_{20}\cdot 20$$

This is a kind of a telescoping sum

$$a_1+(2\cdot a_2-a_2)+(3\cdot a_3-2\cdot a_3)+(4\cdot a_4-3\cdot a_4)+\ldots+(20a_{20}-19a_{20})$$

$$=a_1+a_2+a_3+\ldots+a_{20}$$

callculus42
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  • how did you wrote The number of fishes which are caught by fishermen who caught 1 fish is 1⋅$x_1$=$(a_1−a_2)$⋅1? Kindly explain. –  Jun 20 '20 at 06:56
  • @callculis can you please answer my question asked above in comments? –  Jun 21 '20 at 05:40
  • @Ben I don´t get a notice when you do not use my correct username. – callculus42 Dec 16 '20 at 21:02
  • @Ben I´ve looked at some of your questions. I´m wondering that you have accepted so few answers. – callculus42 Dec 16 '20 at 21:05