## Fun with e

Found a fun problem via facebook last week: find a set of whole numbers that add up to 25, and maximize the product of these numbers. It’s such a fun little puzzle that I won’t give away the solution here. But I got interested in another issue that my friend Aaron pointed out: the connection to e as the “best base”. To get some idea about what I’m getting at, let’s stay with 25 and find the set of numbers (not necessarily whole numbers) that add up to 25 and maximize the product. Actually the numbers have to all be the same — if any two numbers were different, then you could move them closer together and thereby increase their product. So basically what we’re looking at is to maximize the expression $(\frac{25}{n})^n$ for positive integer $n$.

I hope you’ll play around with this REALLY COOL GEOGEBRA APPLET (screenshot below) before continuing.

The thing that really interested me in this problem was how changing the value of A (originally 25) would change the optimal value of n. Sure, for 25 we see that the best value is n=9, but apparently for some larger value of A, then we need n=10. What is the cutoff? When do we switch from 9 to 10?

In fact, you can try it: increase A gradually until the line for n=9 is as high as the line for n=10. You should find yourself at A=25.81 or so. To find the exact value we can do a little algebra. Suppose we have A such that

$(\frac{A}{9})^9 = (\frac{A}{10})^{10}$.

We can do some rearranging…

$A=\frac{10^{10}}{9^9}$

Et voilà! A similar calculation will show that when $A=\frac{9^9}{8^8}$, both 8 and 9 groups give the same product.

Another issue implied by the applet is the link to e. Apparently, we would like the ratio $A/n$ to be close to e to get the maximal product. It’s no coincidence that our original value of 25 is close to 9e (about 24.46). This leads to the interesting inequality

$\frac{9^9}{8^8} < 9e < \frac{10^{10}}{9^9}$.

And based on the context, this is the range where we would choose to divide A by 9, and 9e should be more or less in the middle of this range.

Of course, there’s no reason not to extend this a little on both sides…

$\ldots < 7e < \frac{8^8}{7^7} < 8e < \frac{9^9}{8^8} < 9e < \frac{10^{10}}{9^9} < 10e < \frac{11^{11}}{10^{10}} < 11e < \ldots$

Just go ahead and generalize that…

$\ldots < (n-1)e < \frac{n^n}{(n-1)^{(n-1)}} < ne < \frac{(n+1)^{(n+1)}}{n^n} < (n+1)e < \ldots$

Turn your eyes to that last piece on the right.

$\frac{(n+1)^{(n+1)}}{n^n} < (n+1)e$

We can simplify a little…

$\frac{(n+1)^n}{n^n} < e$

$(\frac{n+1}{n})^n < e$

$(1+\frac{1}{n})^n < e$

Where have I seen that before?