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 or send me an email.
From Jason Armstrong comes a curious conundrum of calendars:
Two friends of Jason, who happen to be married to each other, have birthdays on Feb. 9 and Nov. 18. When written numerically in MM/DD formatting, these dates are 02/09 and 11/18. Jason noted that the latter date includes both the sum and the product of the values in the former date. In other words, 11 = 02 + 09 and 18 = 02 × 09.
How many pairs of dates are there such that one of the dates includes both the product and the sum of the values in the other date (in either order)? Also, note that the order of the dates in the pair doesn’t matter, so “02/09 and 11/18” should be considered the same as “11/18 and 02/09.”
From Graydon Snider comes a dilemma of delivery:
A restaurant at the center of Riddler City is testing an airborne drone delivery service against their existing fleet of scooters. The restaurant is at the center of a large Manhattan-like array of square city blocks, which the scooter must follow.
Both vehicles travel at the same speed, which means drones can make more deliveries per unit time. Assume that (1) Riddler City is circular in shape, as you may recall (2) deliveries are made to random locations throughout the city and (3) the city is much, much larger than its individual blocks.
In a given amount of time, what is the expected ratio between the number of deliveries a drone can make to the number of deliveries a scooter can make?
Extra credit: In addition to traveling parallel to the city blocks, suppose scooters can also move diagonally from one corner of a block to the opposite corner of the block. Now, what is the new expected ratio between the number of deliveries a drone can make and the number of deliveries a scooter can make?
Solution to the last Riddler Express
Congratulations to 👏 Jim Lahey 👏 of Issaquah, Washington, winner of the last Riddler Express.
Last time, you and a friend were shooting some hoops at your local basketball court when she issued a challenge: She would name a number, which we’ll call N. Your goal was to score exactly N points in as many ways as possible using only 2-point and 3-point shots. The order of your shots did not matter.
For example, there were two ways you could score N = 8 points: four 2-pointers or two 3-pointers and one 2-pointer.
Your apparently sadistic friend chose 60 for the value of N. You tried to negotiate this number down, but to no avail. However, she said you were welcome to pick an even larger value of N. Did there exist an integer N greater than 60 such that there were fewer ways to score N points than there were ways to score 60 points?
At first, this all seemed rather counterintuitive. As N increased, the number of ways to score N, which we’ll call the function f(N), should also have increased. However, f(N) was not necessarily a monotonically increasing function.
Let’s see why that was, using smaller values of N. There were zero ways to reach N = 1. There was one way to reach N = 2 (one 2-pointer), 3 (one 3-pointer), 4 (two 2-pointers) and 5 (one 2-pointer and one 3-pointer, in either order). There were two ways to reach N = 6: three 2-pointers or two 3-pointers. Up to this point, as N increased, f(N) never decreased. But there was only one way to reach N = 7: two 2-pointers and one 3-pointer. In function notation, f(7) < f(6). Sure enough, f(N) was not monotonically increasing.
A similar dip occurred around when N was 60. To score 60 points, you could have made 30 2-pointers and no 3-pointers. At the other extreme, you could have made no 2-pointers and 20 3-pointers. Other combinations were possible, swapping three 2-pointers at a time for two 3-pointers. In the end, the number of 3-pointers had to be an even number between 0 and 20 (inclusive), which meant f(60) was 11.
When N was 61, you could have made 29 2-pointers and one 3-pointer. At the other extreme, you could have made two 2-pointers and 19 3-pointers. This time, the number of 3-pointers had to be an odd number between 1 and 19 (inclusive), which meant f(61) was 10. Because f(61) < f(60), 61 points was your desired target.
It turned out that f(N) always decreased when N went from being a multiple of 6 to one more than a multiple of 6. Solvers like Matthew Tanzy and Aidan Dunkelberg were able to prove this by finding an explicit formula for f(N). Matthew found that when N was even, f(N) was equal to floor(N/6)+1; when N was odd, f(N) was equal to floor(N/6+0.5). Sure enough, this confirmed the result that f(61) was less than f(60).
Solution to the last Riddler Classic
Congratulations to 👏 Jared Schmitthenner 👏 of Madison, Wisconsin, winner of the last Riddler Classic.
Last time, the astronomers of Planet Xiddler were back in action! They had used their telescopes to spot an armada of hostile alien warships on a direct course for Xiddler. The armada was scheduled to arrive in exactly 100 days. (Recall that, like Earth, there are 24 hours in a Xiddler day.)
Fortunately, Xiddler’s engineers had just completed construction of the planet’s first assembler, which was capable of producing any object. An assembler could be used to build a space fighter to defend the planet, which took one hour to produce. An assembler could also be used to build another assembler (which, in turn, could build other space fighters or assemblers). However, building an assembler was more time-consuming, requiring six whole days. Also, you could not use multiple assemblers to build one space fighter or assembler in a shorter period of time.
What was the greatest number of space fighters the Xiddlerian fleet could have when the alien armada arrived?
One strategy was to spend the entire 100 days building fighters. There are 2,400 hours in 100 days, so you would have had 2,400 fighters in the end. Alternatively, you could have spent the first six days building an assembler and then the last 94 days making fighters. There are 2,256 hours in 94 days, and with two assemblers that would have resulted in 4,512 fighters.
At this point, you might see how a good strategy was to invest in assemblers early on and then switch over to fighters at the end. After all, having assemblers build more assemblers resulted in exponential growth, whereas building fighters was just linear growth. And sooner or later, exponential growth always blows linear growth out of the water.
Taking this to the extreme, you could have gone through 16 cycles of building assemblers over the first 96 days. This would have resulted in 216, or 65,536, assemblers. Over the final four days, or 96 hours, they would have collectively made 6,291,456 fighters. That’s a whole lot of fighters!
But it was possible to do even better. Working backward, solver Jen McTeague suggested starting with 15 (rather than 16) cycles of building assemblers over the first 90 days. With this strategy, you had half as many assemblers at the end. However, you also had more than twice as many days remaining in which to build fighters (i.e. 10 vs. four), meaning you’d net more fighters.
After those 15 cycles, you had 215, or 32,768 assemblers. Over the final 10 days, or 240 hours, they would have collectively made 7,864,320 fighters. And that was the largest the Xiddlerian fleet could have possibly been.
But was it enough to fend off the alien armada? Find out next time on … The Xiddler!2
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.