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 next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
Riddler Express
From Tom Hanrahan, a maze you can solve without getting lost in a field of corn:
The number in each box tells you how many spaces up, down, left or right you must move. (No diagonal moves, people.) Starting at the yellow six in the bottom left corner, can you make your way to the asterisk?
Riddler Classic
From Michael Branicky, a challenge of currency conversion:
Riddler Nation has two coins: the Dio, worth $538, and the Phantus, worth $19. When visiting on vacation, Riddler National Bank will gladly convert your dollars into Dios and Phanti. For example, if you were to give a bank teller $614, they’d return to you one Dio and four Phanti, since 614 = 1 × 538 + 4 × 19. But if you tried to exchange one dollar more (i.e., $615), then alas, there is no combination of Dios and Phanti the teller could give you, and you won’t get your money’s worth in local currency.
To make the bank teller’s job (and your vacation) as miserable as possible, you decide to bring the largest dollar amount that cannot be converted into Riddler currency. How much money are we talking here? That is, what’s the largest whole number that cannot be expressed as a sum of 19s and 538s?
Extra Credit: Word is that Riddler Nation is considering minting a third currency, worth $101. If they do, then what would be the largest dollar amount that cannot be converted into Riddler currency?
Solution to last week’s Riddler Express
Congratulations to 👏Brennan Carmody 👏 of Atlanta, Georgia, winner of last week’s Riddler Express.
Last week, there was an auditorium with 200 seats, numbered from 1 to 200 and filled to capacity. A speaker overlooking the audience said: “I’m thinking of a rather large whole number. Every seat number in this auditorium evenly divides my number, except for two of them — and those two seats happen to be next to each other.” Which two numbers was the speaker referring to?
Solver Jenny Mitchell made a key realization: Since every seat number, other than the two excluded numbers, had to evenly divide the large mystery number, those two excluded numbers both had to be the largest powers of primes smaller than the total number of seats (which was 200). And because the two seats were consecutive, that meant that one of them had to be even, and therefore a power of 2. The largest power of 2 that’s smaller than 200 is 27, or 128. That must be one of the two seat numbers.
Since the two seats had to be consecutive, you next had to look on either side of 128. Is 129 the largest power of a prime number less than 200? No, since 129 equals 3×43. What about 127? The number 127 is itself prime and can be written as 1271 — so yes, it’s the largest power of a prime less than 200! The two seat numbers were indeed 127 and 128.
Several solvers extended the original problem over Twitter, wondering what happens when we change the size of the auditorium. Interestingly, some auditorium sizes, like 100, have no solution, while a small handful of auditorium sizes, like 13, have two possible solutions — seats 7 and 8, and seats 8 and 9.
And what’s the largest auditorium for which we can (currently) find a solution? That would be one with 282,589,934−3 seats, where the two seat numbers that don’t evenly divide the large number on the speaker’s mind would be 282,589,933 and 282,589,933−1, the largest known Mersenne prime as of this writing. If the auditorium had this many seats, I wonder what number the speaker would have to think of so that only two seats didn’t evenly divide that number … it would have to be really, catastrophically large and … ugh, my brain hurts!
Solution to last week’s Riddler Classic
Congratulations to 👏David Zimmerman 👏 of Los Angeles, California, winner of last week’s Riddler Classic.
Last week, you were asked to navigate your boat around two enemy submarines and safely to a harbor. The subs started out evenly spaced between your boat and the harbor, but you had no further information about their subsequent movement. The subs had to be directly underneath your ship to sink it, and they could track your moves with precision and respond efficiently. How much faster than the sub did your ship have to be to guarantee you could safely avoid them?
As explored in a previous Riddler, if there were just one sub, your ship needed to be about 2.33 times faster than it. The animation below illustrates your ship’s journey, which can be broken down into two parts. First comes a straight line segment that continues until the sub’s expanding radius catches up to you. Then, you wrap around the circumference of the expanding circle until you reach the harbor.
But with two subs, you had to be even faster. This is because the leftmost sub now started closer to the harbor, meaning it’ll reach the harbor in less time — and once a sub reaches the harbor, there’s no way for you to get there safely.
To solve the riddle with two subs, let’s first only consider that leftmost sub. (We’ll come back to the other sub in a minute.) To avoid it, you can employ the same strategy as before: Travel in a line, and then curve around the expanding circumference. David worked out the exact speed with some advanced calculus, finding that you have to be 3.396 times faster than that sub to reach safety. Other solvers turned to their computers, finding a similar result via numerical computation.
But what about that second sub? It turns out the second sub doesn’t affect the answer! Because the subs were evenly spaced at the beginning, the points of tangency — the times when your ship brushed against the sub’s expanding circles — formed similar triangles, guaranteeing that the straight line path that avoided the final sub would narrowly avoid the other sub along the way. So the answer was indeed that your ship had to be 3.396 times faster than the subs.
But what if, instead of two subs, there were N subs evenly spaced between you and the harbor? Just like the case of two subs, it turns out you only have to worry about avoiding the last sub — doing so will guarantee you avoid the other N−1 subs. For example, here’s the optimal trajectory when N equals five:
David went even further, finding an expression for the relative speed v your boat must have to avoid N subs:
\(\sqrt{v^2-1}\exp\left(-\frac{\pi}{2\sqrt{v^2-1}}\right)=N\)
Here’s a graph showing the required speed (relative to the subs) you’d need to escape from anywhere from one to 10 subs:
Finally, David considered what happens when N gets really big — that is, when there are many, many subs. In this case, you’d have to sail very quickly, almost a straight line toward the harbor, finally curving around roughly a quarter of the last growing circle. So as N gets very large, you’re traveling a distance that’s approximately N + 𝜋/2 times longer than the distance between adjacent subs, meaning that’s how many times faster than each sub you’d have to be.
In other words, we’re gonna need a faster boat.
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.