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 Nation has many, many districts, each with the same population. Each district elects one representative (from one of the two major political parties) to the Riddler Nation Legislative Body by majority vote within the district.
However, after recent decisions by the highest court in the land, the redistricting process has resulted in districts that appear to favor one party over the other.
What is the greatest percentage of Riddler Nation’s total population that can vote for one of the parties while resulting in that party not having a majority of seats in the Legislative Body?
From Ivor Traber comes a division of the Lower 48:
Previously on The Riddler, you were tasked with splitting the contiguous United States into two groups of states, themselves contiguous, that were approximately equal in area. This week’s Classic is similar.
The states of California, Colorado, New Mexico and New York have banded together to form the United States of Yellow (USY). Meanwhile, Florida, Texas, Utah and Wyoming are the first states to join the United States of Purple (USP). The remaining 40 contiguous states join either the USY or the USP, such that both new nations have 24 states and these sets of states are themselves contiguous.
Virginia is leaning toward joining the United States of Yellow. Is there any way for it to do this?
(Note: For the purposes of this riddle, Colorado does not border Arizona, and Utah does not border New Mexico. Also, Michigan’s two peninsulas can be thought of as connected, so that Michigan borders Ohio, Indiana and Wisconsin.)
Solution to last week’s Riddler Express
Congratulations to ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â Brian Wilmarth ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â of Silver Spring, Maryland, winner of last week’s Riddler Express.
Last week, you had one year (i.e., 365 days) to complete a task that was easily divisible into portions. Also, you always knew exactly what fraction of the task was left.
Your initial plan was to do exactly 1/365 of the task every day. But then you thought to yourself, “If I did 2/365 today, I could do a little less every day from now on.” And so, on the first day, you completed 2/365 of the task. The next morning, 363/365 of the task remained with 364 days left to complete it, so you did another 2/364 × 363/365 of the task.
If you continued in this fashion every day, dividing the remaining work by the remaining number of days in the year and doing twice that amount, when did you finish?
It turned out that calculating the amount of work you did each day was a tall order. You did 1/365 of the work on the first day and another (2 × 363)/(364 × 365) on the second day. That meant 13,177/13,286 still remained. You had 363 days left to complete it, so on the third day you did 2/363 × 13,177/13,286, or 13,177/2,411,409 of the task.
Some solvers continued figuring out how much of the task you would do each day to figure out when it was done. But, as noted by the team of solvers collectively known as the “MassMutual Riddlers” of Springfield, Massachusetts, there was a simpler way to find the answer — a way that required no calculation at all.
On any given day, you had some amount of the task remaining. You divided it up into equal parts depending on how many days were left, and then you doubled it. On most days of the year, that meant there was still some work left over, and the process continued. But that wasn’t the case when there were just two days left. However much of the task remained at that point, you split it in half (again, because there were two days left) and then doubled it. In other words, you did all the remaining work, leaving nothing to do but lounge around on the final day. In the end, it took you 364 days to complete the task.
As I said, some solvers were determined to figure out precisely how much of the task was done each day. But I have to hand it to this week’s winner, Brian, who plotted how much work remained using a graphing calculator:
From the graph, you can see that work on the task was definitely frontloaded, even if you only finished a day ahead of schedule in the end.
Solution to last week’s Riddler Classic
Congratulations to ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â Tom Knief ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â of Des Moines, Iowa, winner of last week’s Riddler Classic.
Last week, you and a friend climbed two of the tallest towers on Planet Xiddler, which happened to be in neighboring cities. You both traveled 100 meters up each tower on a clear day. Due to the curvature of the planet, you could barely make each other out.
Next, your friend returned to the ground floor of their tower. How high up your tower did you have to be so that you could just barely make out your friend again?
There wasn’t a lot of information to work with in this problem. However, there were a few additional reasonable assumptions you could make. First was that the planet was spherical. Second was that both towers started at the same altitude (although, technically, this was covered by the assumption that the planet was a perfect sphere). And while no height was given for you or your friend, you could assume that you were both much shorter than 100 meters, so that height was not a relevant part of the problem. And finally, since the towers were in “neighboring cities,” you could assume that the radius of the planet was much larger than the heights of the towers. (We’ll return to that last assumption, however.)
With these assumptions, how far can you see if you are a height h off the ground on a planet with radius R? If your line of sight has a distance d, then with an assist from the Pythagorean theorem (shown below) you know that d2 = (R+h)2 − R2. Canceling out the R2 terms, that means d2 = 2Rh+h2. And because we assumed that the radius R was much greater than the height h, you could ignore the h2 term in that last expression, since it was so much smaller than the 2Rh term. That gave you d = √(2Rh).
Now, when you and your friend could barely see each other when you were both 100 meters up your respective towers, that meant you each had the same line of sight — you both saw to the point on the ground that was halfway between the towers. To see your friend when they returned to the ground floor, your line of sight had to double. In the equation d = √(2Rh), when the value of d doubles, the value of h has to quadruple in order maintain balance. And that meant your new height was 400 meters.
While 400 meters was the answer I was looking for, it was still an approximation for when the radius of the planet was much larger than 100 meters. As Josh Silverman showed, this was a very reasonable approximation, since a rocky planet had to have a radius on the order of at least 100,000 meters to even be spherical. Several solvers, like Emma Knight and Dean Ballard gave more precise expressions for how the height varied with radius. Laurent Lessard made a plot, which you and your friend could conceivably have used to measure the radius of the Xiddler (although making such a measurement would have been very tricky):
If Xiddler were close in size to Earth and your measurements were off even by just a few micrometers, your estimate for Xiddler’s radius could have been off by a factor of two or more!
Want more puzzles?
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 firstname.lastname@example.org.