Ever so equable III

Let the area of the base of the prism be $B$ , let the perimeter of the base be $P$ , let the number of sides of the base be $n$ , and let the length of the lateral sides be $h$ . Then the volume is $V = Bh$ , the surface area is $S = 2B + Ph$ , and the total edge length is $E = 2P + nh$ .

Setting $S = E$ and solving for $h$ gives $h = \frac{2(B - P)}{n - P}$ , and then setting $V = E$ and substituting $h$ and solving for $B$ gives $\frac{1}{2}(P + n \pm \sqrt{(n - P)(n + 3P)})$ .

For $B$ to be a real number, $(n - P)(n + 3P) > 0$ or $n > P$ . If $n > P$ and $h = \frac{2(B - P)}{n - P}$ , then for $h$ to be a positive number, $B > P$ .

A polygon with $n$ sides and a fixed perimeter will have its largest area when it is a regular polygon, and since $n > P$ , the largest side of that regular polygon will be $1$ . Since the area of a regular polygon with $n$ unit sides is $\frac{n}{4 \tan (\frac{\pi}{n})}$ , we have $B < \frac{n}{4 \tan (\frac{\pi}{n})} < n = P$ for $n \leq 12$ .

However, $B > P$ , so we must have $\frac{n}{4 \tan (\frac{\pi}{n})} > n$ , or $\tan (\frac{\pi}{n}) < \frac{1}{4}$ , which is not true until $n \geq 13$ .

Therefore, the smallest $n$ such that a right $n$ -gonal prism exists whose volume, surface area and total edge length are all numerically equal is $\boxed{n = 13}$ .