Caesar Mathematics : I

Let the first pile initially have $x_1$ logs, the second pile $x_2$ , etc.; the last ( $n$ th) pile will thus have $x_n$ .

The first pile is in a special position because $\frac{1}{n}$ logs are initially withdrawn from it, and at the end of the process is it replenished in from the $n$ th pile, whereas all other piles are first filled with logs from the previous piles, and only then have $\frac{1}{n}$ th of the logs withdrawn from them.

So, let us observe any pile but the first pile. Let's call our pile $k$ . Initially, it had $x_k$ amount of logs, and a certain amount of logs were added to it - let's call it $y$ - from the $(k-1)$ th pile. Then, from the overall sum of logs of $y+x_k$ logs was withdrawn a $\frac{1}{n}$ th part. So we're left with $(y+x_k)\frac{n-1}{n}$ logs (in that pile). Thus,

$(y+x_k)\frac{n-1}{n}=A. \rightarrow (1)$

In the previous - $(k-1)$ th pile, if it wasn't the first pile (i.e. if $k \neq 2$ ), it must've had $A$ logs. (The first pile only acquires $A$ amount of logs after it is replenished from the $n$ th pile). Therefore, before the transfer of $y$ logs it held $A+y$ logs. By the condition the withdrawn amount of $y$ logs constitutes $\frac{1}{n}$ th part of $A+y$ , i.e.

$y=\frac{1}{n}(A+y). \rightarrow (2)$

From here we find $y=\frac{1}{n-1}A.$ Substituting this into the first expression, we obtain $x_k=A$ .

Hence, in every pile (maybe with an exception of first and second piles - prior to being excluded from consideration), there were initially $A$ logs:

$x_3=x_4=...=x_n=A. \rightarrow (3)$

The unknown $x_1$ can be found in a following way; by the condition, $x_1$ has $\frac{1}{n}$ th of its logs withdrawn from it. We're left with $x_1\frac{n-1}{n}$ logs. At the end of the process, a certain amount of logs - $y$ - is added to it from the last pile. Hence,

$y+x_1\frac{n-1}{n}=A. \rightarrow (4)$

Applying the same reasoning to the $n$ th pile, we find that $y=\frac{1}{n-1}A$ . Inserting this into expression $(4)$ we obtain

$x_1=\frac{(n-2)n}{(n-1)^2}A. \rightarrow (5)$

To investigate for $x_2$ we dig up

$\left(\frac{1}{n}x_1+x_2 \right)\frac{n-1}{n}=A, \rightarrow (6)$

where $x_1$ is determined by expression $(5)$ . Solving this equation, we obtain

$x_2=\frac{n(n-1)-(n-2)}{(n-1)^2}A.$

And thus,

$\boxed{x_1=\frac{n^2-2n}{(n-1)^2}A};$ $\boxed{x_2=\frac{n^2-2n+2}{(n-1)^2}A;}$ $\boxed{x_3=x_4=...=x_n=A.}$

CAESAR

"Caesar and Maurice agreed to divide the resources into 101 piles" All piles have the same amount of logs.

"If, before shifting, the amount of logs in plies 1, 2, and 3 combined is 150..." Piles 1, 2, and 3 have a combined number of 150 logs.

Each pile has $\frac{150}{3}=50$ , so 101 piles would be $50*101=5050$