Can You Break Max’s Wand?

This week’s Riddler Express asks:

Max the Mathemagician is calling for volunteers. He has a magic wand of length 10 that can be broken anywhere along its length (fractional and decimal lengths are allowed). After the volunteer chooses these breakpoints, Max will multiply the lengths of the resulting pieces. For example, if they break the wand near its midpoint and nowhere else, the resulting product is 5×5, or 25. If the product is the largest possible, they will win a free backstage pass to his next show. (Amazing, right?)

You raise your hand to volunteer, and you and Max briefly make eye contact. As he calls you up to the stage, you know you have this in the bag. What is the maximum product you can achieve?

Extra credit: Zax the Mathemagician (no relation to Max) has the same routine in his show, only the wand has a length of 100. What is the maximum product now?


It’s pretty easy to prove that once you choose the number of times to break the wand, the product-maximizing approach to the size of the segments is to have equally-sized sections. Understanding this, to win the game, we have to figure out how many pieces we should end up with.

Evaluating each number of segments, we see that for a 10-unit long wand, the maximum product occurs at 4 segments (three divisions), with each segment having a length of 2.5 units. This gives a total product of 39.0625.

Extra Credit: With a 100-unit long wand, we can use the same process to identify that there should be 37 sections (36 divisions) of length 2.703 units. This gives a total product of (about) 94,740,61,716,781,810.

Extra Extra Credit: But wait; there’s more! Why 4 for the 10-unit and 37 for the 100? If we look at the lengths of the segments, we see that we’re approaching Euler’s number — e. Zach Wissner-Gross, Laurent Lessard, and Ollie Roeder probably know why this is the case, but I don’t. I look forward to finding out!