Newsletter

Rick, United States

Monday, January 5, 2026

"Rather than deal with fractions, I decided to use excel and a starting length of fence equal to 12^7 so all math would result in integers. The following table has four columns

🛑 Month
🛑 Unpainted fence Length. This is calculated for each row by subtracting the previous month’s unpainted length from the amount painted during that month. The initial length is set to 12^7 for January. This will be sufficient to ensure all calculations will have integer results so no rounding will occur.
🛑 Amount Painted. Equals the Unpainted fence Length times the Month all divided by 12.
🛑 Difference. The difference between what was painted between the current month and the previous month.

Positive numbers represent an increase in amount painted and negative numbers indicate less amount painted. Zero indicates equal parts painted between two months.



From the table, we can observe that for months 3 and 4, each resulted in in the same length of fence painted (6842880) which was the greatest length compared to other months. From a purely mathematical point of view, the most paint was used in both March and April. From a physics point of view, if the fence was in the northern hemisphere, March is colder than April, so the viscosity of the paint would have been thicker, and more paint would have been required in March. "

Kalmakka, Reddit

Thursday, January 8, 2026

"A pattern seems to be that if \(x\) is square, then you do the most work in month \(\sqrt{x}\).

Let us say \(x=n^2\). At the start of month \(n\), \(p\) meters of fence remains.

That means that in month \(n\) we paint \(p\cdot\left(\frac{n}{n^2}\right)=p\cdot\left(\frac{1}{n}\right)\) meters (leaving \(\frac{p(n-1)}{n}\) unpainted).

In month \(n+1\) we paint:
$$ \left(\frac{p(n-1)}{n}\right)\left(\frac{n+1}{n^2}\right) =\frac{p(n+1)(n-1)}{n^3} =\frac{p(n^2-1)}{n^3} <\frac{p}{n}. $$

Since we painted \(\frac{n-1}{n^2}\) portion of the unpainted fence in month \(n-1\), leaving a portion of \(\frac{n^2-n-1}{n^2}\) unpainted, and that this remaining portion equals \(p\) meters, we know that the amount we painted in month \(n-1\) is:
$$ p\cdot\left(\frac{n-1}{n^2-n-1}\right) < p\cdot\left(\frac{n-1}{n^2-n}\right) = p\cdot\left(\frac{1}{n}\right). $$

So we see that in month \(n\) we paint more than in both month \(n-1\) and in month \(n+1\).

If we reason that the function is concave, then this local maximum is also the global maximum.

Whenever \(x=n(n+1)\) we have two maximums, at \(n\) and \(n+1\). E.g. for \(12\) we have that in month \(3\) we paint \(\frac{1}{4}\) of whatever remains, leaving \(\frac{3}{4}\) of that, and in month \(4\) we paint \(\frac{1}{3}\) of that \(\frac{3}{4}\). Since \(\frac{1}{4}=\left(\frac{3}{4}\right)\cdot\left(\frac{1}{3}\right)\), we paint the same amount in these two months. "

Edam_Cheese, Reddit

Thursday, January 8, 2026

"A bit of algebraic manipulation gives the amount we paint on the \(n\)th month as \(n\cdot 12^{-n}\cdot(12-1)(12-2)\cdots(12-(n-1))\). I couldn't get the derivatives to work nicely enough to find the maxima from this, so evaluating it for each \(n\) still seems the best way to go.

For the same problem over \(x\) months: I assume that on the \(n\)th month we'll paint \(\frac{n}{x}\) of what remains. So for \(13\) months, we paint \(\frac{1}{13}\)th in Jan, then \(\frac{2}{13}\)th of what's left in Feb, and so on. We can simply swap \(12\) in the above formula for \(x\) to get the expression for the amount we paint on the \(n\)th month, given we take \(x\) months total.

This last part is from inspection of the data, but I can't prove it. For the problem of painting a fence over \(x\) months, we do the most painting on the \(\left\lceil \frac{1}{2}\sqrt{1+4x}-\frac{1}{2}\right\rceil\)th month. When the expression is an integer before taking the ceiling, we also do the same amount of painting on the following month. "

Vidarino, Reddit

Thursday, January 8, 2026

"I poked at this for a bit, and found an algebraic solution, but it's not super-pretty. I think it's correct, though, and should be pretty generalizable. No idea about the source of the problem.

Let \(R_n\) be the amount of work that remains at step \(n\).

It can be expressed as the product of \(\left(1-\frac{k}{12}\right)\) for each \(k\) in \([1,n-1]\). (This can be "algebra-notated" using the product function (capital PI)

Let \(W_n\) be the work for the month \(n\): \(\frac{n}{12}\cdot R_n\).

Since we're only interested in finding out the maximum work, let's compare successive months. This isn't very rigorous, but let's assume that the work done each month first increases and then decreases. To find out when this happens we can solve: \(\frac{W_{n+1}}{W_n}>1\). (I.e. when ratio falls below \(1\) the amount of work starts to decrease.)

Expanding this into all its glory is again quite unsuitable for a Reddit comment field, I'm afraid, but we get a bunch of cancellations, and are left with something that can be simplified to \(\frac{(n+1)(12-n)}{12n}>1\).

Expand and simplify more: \(n^2+n-12<0\).

Factorise: \((n+4)(n-3)<0\).

So by inspection we see that the work ratio increases up until \(n=3\) (at which we can see that it's actually \(0\), so the amount of work will be the same at \(n=4\)).

So, there we go. Hardly elegant, but it is algebraic rather than brute force. "

CyberMonkey314, Reddit

Friday, January 9, 2026

"There's no need for an explicit formula for the amount used in each month to answer the question as stated, or for any differentiation.

Say at the beginning of month \(n\), you have a fraction \(R_n\) of the fence remaining to be painted (so \(R_1=1\), the whole fence). You then paint \(p_n=\frac{nR_n}{12}\) of the fence, so \(R_{n+1}=R_n\left(\frac{12-n}{12}\right) \).

In the next month, you paint \(p_{n+1}=\frac{(n+1)R_{n+1}}{12}\) of the fence. Now, the "trick" is to think about the ratio of the amounts of paint used month on month; i.e. \(\frac{p_{n+1}}{p_n}\).

Substituting in, this is \(\frac{p_{n+1}}{p_n}=\frac{(n+1)(12-n)}{12n}=\frac{1}{n}+\frac{11-n}{12}\).

Why does this help? When the ratio is greater than \(1\), you've used more paint in month \(n+1\) than in month \(n\). When it's less than \(1\), you've used less. When it is \(1\), you use the same amount. So the maximum amount of paint is used either in the first month \(n\) when \(\frac{p_{n+1}}{p_n}<1\), or it's a tie between months \(n\) and \(n+1\) if \(\frac{p_{n+1}}{p_n}\) is ever exactly \(1\).

Neatly, the quantity \(\frac{1}{n}+\frac{11-n}{12}\) is strictly decreasing! It starts above \(1\) and ends below \(1\). Solving \(\frac{p_{n+1}}{p_n}=1\) leads to \((n+1)(12-n)=12n\), whence \(n=3\) (we ignore the negative solution \(n=-4\)).

Since this solution is an integer, there are two months in which the maximum amount of paint is used, i.e. month \(3\) and month \(4\) (had it not been an integer, a single month would be maximal).

I hope that makes sense (and doesn't have any typos) "