, 2023

Since $64 = 2^6$, its positive
divisors are 1, 2, 4, 8, 16, 32, and 64.

The sum of these divisors is $1 + 2 + 4 + 8 +
16 + 32 + 64 = 127$.
Suppose that the four consecutive integers that Fionn originally
wrote on the blackboard were $x$,
$x+1$, $x+2$, and $x+3$.

When Lexi erases one of these integers, the sum of the remaining three
integers is equal to one of the following: $$\begin{align*} (x+1) + (x+2) + (x+3) & = 3x + 6 \\ x + (x+2) + (x+3) & = 3x + 5 \\ x + (x+1) + (x+3) & = 3x + 4 \\ x + (x+1) + (x+2) & = 3x + 3\end{align*}$$ We are told that
the sum of these integers is 847.

We note that $847 = 3 \cdot 282 + 1$,
which is one more than a multiple of 3. Since $3x+3$ and $3x+6$ are always multiples of 3 and $3x+5$ is 2 more than a multiple of 3, then
we must have $3x + 4 = 847$ and so
$3x = 843$ or $x = 281$. (Alternatively, we could have
set each of the four sums above equal to 847 to determine in which case
or cases we obtained an integer solution for $x$.)

Therefore, the original integers were 281, 282, 283, 284 and Lexi erased
$x + 2 = 283$.
From the given information, the 7 terms in the
arithmetic sequence are $$d^2,~~d^2 + d,~~d^2 + 2d,~~d^2 + 3d,~~d^2 + 4d,~~d^2 + 5d,~~d^2 + 6d$$ Since the sum
of these 7 terms is 756, we obtain the following equivalent equations:
$$\begin{align*} d^2 + (d^2 + d) + (d^2 + 2d) + (d^2 + 3d) + (d^2 + 4d) + (d^2 + 5d) + (d^2 + 6d) & = 756 \\ 7d^2 + 21d & = 756 \\ d^2 + 3d & = 108 \\ d^2 + 3d - 108 & = 0 \\ (d + 12)(d - 9) & = 0\end{align*}$$ and so $d = -12$ or $d
= 9$.

The corresponding arithmetic sequences are $$144, 132, 120, 108, 96, 84, 72 ~~\text{ and } ~~ 81, 90, 99, 108, 117, 126, 135$$