Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
Riddler Express
From Dave Moran comes a riddle so foul it’s sure to bust your bracket:
In NCAA men’s basketball, when a team has committed between six and eight fouls in a half, the opposing team is “in the bonus.” This means that the next time they are fouled while not in the act of shooting, they get to shoot at least one free throw. If the player makes that first free throw, they get to shoot a second.2 If either free throw is missed,3 one team will rebound the ball, and play will continue with the rebounding team in possession.4
Assume that an average team scores exactly 1 point per offensive possession, a figure that accounts for multiple shots if the team rebounds its own miss (or misses) on a single trip, and that it rebounds 15 percent of its own missed free throws.5
Now suppose you are the coach of a team playing an average opponent that’s in the bonus. The other team has the ball, the game is tight, and you want to minimize the expected number of points your opponent will earn on this particular possession. How low does the ball-handler’s free throw shooting percentage need to be for you to instruct your team to foul that player (when they are not in the act of shooting)?
The solution to this Riddler Express can be found in the following column.
Riddler Classic
The rules for men’s basketball in the Riddler Collegiate Athletic Association’s (RCAA) are a little different from those in the NCAA. In the NCAA, when a player is fouled attempting a 3-point shot and misses, they always get three free throw attempts, regardless of how many fouls the opposing team has committed.
But in the RCAA, a player must earn each additional foul shot by making the previous one (similar to the “bonus” rules of the NCAA mentioned in the Express). In other words, a player can take a second shot if they make the first, and they can take a third shot if they make the second.
Suppose a player on your team has a known shooting profile: Their probability of making the first free throw is p, their probability of making the second is q, and their probability of making the third is r, such that no two of these probabilities are equal. Meanwhile, their expected number of points made for any given three-point foul (which can be computed from p, q and r) is also known.
What is the greatest number of distinct shooting profiles that are made up of these three different probabilities — p, q and r, in some order for the three shots — that can result in the same overall expected number of points?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏Matthew Depew 👏 of Plano, Texas, winner of last week’s Riddler Express.
Last week, you had to solve a felt-tip geometric puzzle illustrated by the one and only Catriona Agg:
The two larger squares were congruent, and the smaller square made a 45 degree angle with one of the larger squares. Both larger squares touched the circle at one corner, while the smaller square touched the circle at two corners.
How many times greater was the area of one of the larger squares than the area of the smaller square? If we call the side length of a larger square x and the smaller square y, then you had to find the ratio x2/y2.
While there were dozens of solutions that leveraged trigonometry and the geometry of circles, I’d like to highlight a particularly creative solution from Michael DeLyser of State College, Pennsylvania. Michael drew two additional chords on top of Catriona’s felt-tip sketch and also copied one quarter of the smaller square (i.e., an isosceles right triangle), pasting it to the right of a larger square:
The vertical chord is twice the side length of one of the larger squares, or 2x. But what about the horizontal chord? Because of the symmetry about that isosceles right triangle in the top right of the figure, the two chords had to be equal in length. But the horizontal chord was also x plus one-and-a-half times the diagonal of the smaller square, or 3y/√2.
That allowed you to set up the following equality: 2x = x + 3y/√2, or x = 3y/√2. Dividing both sides by y meant the ratio of the squares’ side lengths, x/y, was 3/√2. Finally, squaring this ratio gave you the ratio of the areas, which was 9/2, or 4.5.
If you enjoyed this puzzle, Catriona has been posting similar (if not better) ones every day for some time, which means about seven of them since this puzzle came out last week, if my math is right. Go check them out!
Solution to last week’s Riddler Classic
Congratulations to 👏David Ding 👏 of Natick, Massachusetts, winner of last week’s Riddler Classic.
Last week, you first saw that for some perfect squares, when you removed the last digit, you got another perfect square. For example, when you removed the last digit from 256 (162), you got 25 (52).
The first few squares for which this happened were 16, 49, 169, 256 and 361. What were the next three squares for which you could have removed the last digit and gotten a different perfect square? How many more could you find? (Bonus points for not looking this up online or writing code to solve it for you!)
Yes, you could have looked it up on OEIS. It turned out this was sequence A023110. But you were expected to do more than look up this sequence or have your computer crunch through a few thousand square numbers, remove the last digit and check to see if the result was still a square.
Instead, you were asked to solve this problem analytically. But how was this possible?
Suppose a2 was a square number in this sequence, meaning its square root was the whole number a. When you removed the last digit, you got another perfect square, which we can call b2. Removing a digit is equivalent to subtracting the value of the last digit and then dividing by 10. As the numbers in this sequence grew, subtracting the last digit did little to change the number’s value when compared to dividing by 10. In other words, a2 was a shade more than 10 times b2, which meant a was a shade more than √10 times b, i.e., the rational number a/b was slightly greater than √10.
So then how could you find squares in this sequence by hand? The answer was to identify rational approximations of √10 that were greater than √10 and then check the squares of the numerator and denominator.
The square root of 10 is approximately 3.16227766017. An efficient way to find its rational approximations is to use continued fractions, which alternate being slightly less and slightly more than the irrational value. You could also start with two rational numbers — one less than √10 and the other greater than √10 — and calculate successive mediants. Without going into too much detail, the mediant of two nearby fractions a/b and c/d is (a+c)/(b+d), a value that will always fall between the two fractions.
Let’s start with 3/1 and 4/1, the latter of which gives the “square pair” 16 and 1 — remove the last digit of 16 and you get 1. Their mediant is 7/2, which gives the square pair 49 and 4. Because 7/2 is still greater than √10, we should next check the mediant of 3/1 and 7/2. This is 10/3, which does not result in a square pair, since removing the last digit of 100 (102) does not give you 9 (32). The next mediants that gave a proper square pair were 13/4 (169 and 16), 16/5 (256 and 25) and 19/6 (361 and 36). In fact, 192 was so close to 10 times 62 that multiplying them both by four and nine gave two more pairs: 382 and 122 (i.e., 1,444 and 144) and 572 and 182 (i.e., 3,249 and 324).
But you still had to find one more term in this sequence. The mediant of 3/1 and 19/6 was 22/7, which was less than √10. That meant the larger square (484) was less than 10 times the smaller (49). While the fraction 22/7 may be a good approximation for 𝜋, it didn’t help you here. The next mediant that was greater than √10 turned out to be 136/43, which gave the pair 18,496 and 1,849.
Continuing with these mediants (and checking their square multiples) will continue producing numbers in this sequence. Meanwhile, some solvers, like Goh Pi Han and Vaidhy Mahalingam, used Pell equations to identify more squares.
For extra credit, you looked for squares such that removing the last digit produced a second square and removing the last two digits produced a third square. You were given one example of such a triple: 169 (132), 16 (42) and 1 (12). Indeed, as Emma Knight proved, that was the only such triple.
Solution to last week’s Bonus Riddler Express
Congratulations to 👏Mary Lo 👏 of Alhambra, California, winner of last week’s Bonus Riddler Express.
Last week, you were caring for an infant and trying to interpret their cry. Were they hungry? Were they tired? Did they need a diaper change?
Suppose you had an infant who napped peacefully for two hours at a time and then woke up, crying, due to hunger. After eating quickly, the infant played alone for another hour and then cried due to tiredness. This cycle repeated several times over the course of a 12-hour day. (Your rock star baby slept peacefully 12 hours through the night.)
You were working in an adjacent room when your partner walked out and handed you the baby monitor. You had completely lost track where in the day this happened. You continued working for another 30 minutes, and you did not hear the baby cry during this time. What was the probability that the next time your baby cried they would be hungry?
This was an extension of the prior week’s Riddler Express, in which you heard the baby cry after exactly 30 minutes. To solve that problem, Alex Vornsand generated the following helpful diagram:
The answer to that prior riddle was 1/2, since there were exactly four moments throughout the day that preceded a “hungry cry” by 30 minutes, and another four moments that preceded a “tired cry” by 30 minutes.
But this time around, no cry came. That meant you had to be in one of the unshaded regions in Alex’s diagram. The periods preceding the “hungry cry” were each three times longer than the corresponding periods preceding the “tired cry,” which meant the answer was 3/(3+1), or 3/4.
Better get that bottle ready!
Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at riddlercolumn@gmail.com.