Add up the eggs

We start by considering a simple example: One chicken that lays 20 eggs. This means that all values, from $a_1$ to $a_{20}$ , are equal to one.

We consider Chandler's guess first, and immediately discover that it is wrong. His formula gives a solution of 210 eggs. Chandler's formula would only be correct if $a_n$ gave the number of chickens that laid exactly $n$ eggs.

Alice and Bob's solutions are both correct here, so we consider them both.

First, Bob's solution. Specifically, Bob's solution is of the form $n(a_n-a_{n+1})$ . Let's consider a single chicken. If the chicken has laid less than $n$ eggs, then obviously they do not contribute to $a_n$ and $a_{n+1}$ , so they will not affect the total difference. If the chicken has laid more than $n$ eggs, $a_n$ and $a_{n+1}$ will both be incremented, so the difference will not be affected. Only if the chicken lays $n$ eggs will the difference $a_n-a_{n+1}$ be incremented , so the term $a_n-a_{n+1}$ describes the number of chickens who have laid exactly $n$ eggs. If we define $b_n$ to be the number of chickens that lay exactly $n$ eggs (noticing here that since no chickens lay more than 20 eggs, $a_{20}=b_{20}$ ), we see Bob's expression simplify to $20b_{20}+19b_{19}+18b_{18}+\dots+2b_{2}+b_{1}$ It is trivial to verify that this will be the total number of eggs laid.

Last, we have Alice's solution. We notice that a chicken that lays $n$ eggs will increment exactly $n$ values, from $a_1$ to $a_n$ . Because of this, we see that this sum will count each egg that each chicken lays. As well as this, we can see that this expression is equivalent to Bob's by replacing each $a_n$ term with $(n+1)a_n-na_n$ and factoring.