Easter Egg Hunt!

Let's say there were $x-1, x, x+1$ eggs in the baskets respectively. Therefore, the total number of eggs would be $3x$ . If $3x$ is between 0 and 300 inclusive, or in the interval $[0,300]$ , there are 101 different ways that this could happen; however, if $x=0$ , $x-1$ would be a negative number, which in this case is a situation that we have to throw out. This leaves our answer to be $\boxed {100}$ .

The "minimum" solution is: (0, 1, 2). Here, this is designated as the 1st solution. The first few solutions may be listed as: (0, 1, 2):[solution 1], (1, 2, 3):[solution 2], (2, 3, 4):[solution 3], and so on ... .Notice that the maximum number in the sequence of the three consecutive numbers is one more than the number of the solution (i.e. for the 3rd solution: 4, the maximum number in the set 2, 3, 4 is one more than 3, the number of the solution). The last possible solution must satisfy: (a - 2) + (a - 1) + a = 300, which is solved when a = 101. The last possible solution is (99, 100, 101). The largest integer in the last solution is 101 which then corresponds to the 100th solution.

The maximum number of eggs that can be collected is 300. The boys collect consecutive amounts of eggs. Labeling the smallest number of eggs collected $n$,

$$ n + (n + 1) + (n + 2) &\leq 300\ \implies 3n+ 3 &\leq 300\ \implies n &\leq 99. $$

The number of eggs collected cannot be negative, so there is 100 possible values for $n$, which gives the answer.