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!

What Does The Crying Baby Want?

This week’s Riddler Express asks:

As any caretaker can tell you, one challenging part of caring for an infant is interpreting their cry. Are they hungry? Are they tired? Do they need a diaper change?

Suppose you have an infant who naps peacefully for two hours at a time and then wakes up, crying, due to hunger. After eating quickly, the infant plays alone for another hour, and then cries due to tiredness. This cycle repeats several times over the course of a 12-hour day. (Your rock star baby sleeps peacefully 12 hours through the night.)

You’re working in an adjacent room when your partner walks out and hands you the baby monitor. You’ve completely lost track where in the day this happens. You continue working for another 30 minutes, then you hear the baby cry. What’s the probability that your baby is hungry?


Another fun, easy, but counter-intuitive question. One might think that since the baby spends twice as much time sleeping as playing, they are twice as likely to be hungry as tired. But, to answer this question it can be helpful to map out the schedule visually:

With this diagram we can see that there are eight possible times that correspond with the time your spouse hands you the baby monitor, and four of those are during the nap phase and four during the play phase. When the baby starts crying, there’s a 50% chance it’s because they need to be fed.

Can You Hunt More Mystery Numbers?

This week’s Riddler Express asks:

By all accounts, Riddler Nation had a lot of fun hunting for the mysterious numbers a few weeks back. So here’s what we’re going to do: For the next four weeks, the Riddler Express will feature a similar puzzle that combines multiplication and logic. We’ll be calling these CrossProducts.

For your first weekly CrossProduct, there are five three-digit numbers — each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, are shown in the bottom row.

Can you find all five three-digit numbers and complete the table?


Like the Riddler Classic from several weeks ago, this question can be solved by finding the factors of numbers given and deducing (through logic or software) what the three-digit numbers are. Rather than write it up again, I’ll direct you to my writeup from before.

This week’s answers are 359, 591, 818, 578, and 572.

Fun With Unit Cubes

This week’s Riddler Express asks:

If you have young children (or if you’re still a child at heart), you probably have small blocks somewhere in your home.

I recently found four cubic blocks in a peculiar arrangement. Three of them were flat on the ground, with their corners touching and enclosing an equilateral triangle. Meanwhile, the fourth cube was above the other three, filling in the gap between them in a surprisingly snug manner. Here’s a photo I took of this arrangement:

If you too have blocks at home (I mean, of course you do), see if you can make the same arrangement.

Now, if each of the four cubes has side length 1, then how far above the ground is the bottommost corner of the cube on top?


The first step I took in solving this problem was to figure out what to measure to arrive at the answer. Consider this isometric diagram of the arrangement:

In this perspective, we’re looking directly through the nearest base unit cube (which appears as a solid-lined square in the middle). The portion of the suspended cube that sits within the recess created by the other three cubes is a pyramid. The base of this pyramid is an equilateral triangle of length 1 (inherited from the sides of the supporting unit cubes). For the next step, I’ll focus on this pyramid alone:

In this view of the pyramid (looking directly down on it), we see the base is a unit equilateral triangle, and the walls are isoceles right triangles with hypotenuse 1 and edges of \(\frac{1}{\sqrt{2}}\). To figure out the total height of the pyramid, we’ll need to inspect the triangle which represents the horizontal projection of this pyramid.

This triangle, depicted below, has a base of \(\frac{\sqrt{3}}{2}\), and one of the sides is the \(\frac{1}{\sqrt{2}}\) vertex of the pyramid.

The height of this triangle, we can figure out using trigonometry, is \(\frac{1}{\sqrt{6}}\). This number tells us how deep into the recess the suspended cube falls. From here, we can simply subtract this from 1 to get our answer: \(1-\frac{1}{\sqrt{6}}\approx\)0.592.