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How Often Does The Senate Vote In Palindromes?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and 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 Robbie Hart, a present-day parliamentary procedure puzzle:

On Monday, the Senate voted 81-18 to end the government shutdown. This naturally grabbed the Riddler’s attention: It’s a palindrome! The vote tally reads the same forward and backward. This specific tally was made possible by the absence of John McCain. But do senators need to be absent to create palindrome tallies? If so, what numbers of absences will do the trick?

Extra credit: How many palindromic Senate votes have occurred in the past three decades?

Submit your answer

Riddler Classic

From Josh Streeter, graph theory meets ancient Greece in a puzzle that takes two classic mathematical ideas and mashes them together:

The famous four-color theorem states, essentially, that you can color in the regions of any map using at most four colors in such a way that no neighboring regions share a color. A computer-based proof of the theorem was offered in 1976.

Some 2,200 years earlier, the legendary Greek mathematician Archimedes described something called an Ostomachion. It’s a group of pieces, similar to tangrams, that divides a 12-by-12 square into 14 regions. The object is to rearrange the pieces into interesting shapes, such as a Tyrannosaurus rex. It’s often called the oldest known mathematical puzzle.

Your challenge today: Color in the regions of the Ostomachion square with four colors such that each color shades an equal area. (That is, each color needs to shade 36 square units.)

Extra credit: How many solutions to this challenge are there?

Submit your answer

Last week we presented two puzzles from this year’s MIT Mystery Hunt. The solutions to both were English words or phrases, and they are explained below. Full disclosure: These were hard! But hey, if hunting mysteries were easy, everybody would do it.

Solution to last week’s Puzzle 1

Congratulations to рџ‘Џ Avery Ellis and Matt Hurley рџ‘Џ of Chicago, winners of last week’s Puzzle 1!

The first puzzle last week from the MIT Mystery Hunt, titled “Studies in two-factor authentication” and authored by Brandon Avila, presented you with some mysterious text and an even more mysterious, staticky-looking image. The introductory text read: “Ugh! Dad says the computer will hurt my eyes, but I doubt that’s his prime concern. Time to see what requires such complex security.”

This title and text, like all of the “flavor text” at the Mystery Hunt, provided important clues. “Hurt my eyes” suggests that you should look very closely at the image, and perhaps enlarge it, while the words “factor,” “prime” and “complex” suggested that this puzzle has to do with factoring prime numbers in the complex plane. (A complex number is a number that has a real part and an пїЅпїЅпїЅimaginary” part, where the “imaginary” number, i, is equal to the square root of -1.)

The image was a 250-by-250 section of the first quadrant in the complex plane beginning at (0, 0) in the bottom left corner.

Each black dot represents a Gaussian prime (a prime number in the ring of Gaussian integers), and each colored dot represents a composite number with exactly two factors. (A prime number has no factors but itself, and a composite number is any number that is not a prime but can be factored into primes.) For each color — red, orange, yellow, green, blue, purple — the composite numbers are factored, and their factors are connected pairwise in the plane by a line. You can see that in action in a zoomed-in portion of the image below:

Let’s walk carefully through the first example. First, locate all of the red squares in the original image. There are five of them. (They’re not visible in the portion above.) Start counting from the bottom left of the image. One red square is located 43 pixels to the right and 80 pixels up. Another is located 76 pixels to the right and 65 pixels up. The others are 68 and 87; 58 and 103; and 46 and 125.

Since we’re working in the complex plane, the first coordinate is the number’s real part, and the second coordinate is its imaginary part. So we can write the first red square, for example, as 43 + 80i, the second as 76 + 65i, and so on.

The complex number 43 + 80i has exactly one factorization: (8 + 3i)(8 + 7i). We then locate those two points on the plane — (8, 3) and (8, 7) — and connect them with a red line. That turns out to be the left upright of our letter B. 76 + 65i has one factorization, too: (8 + 3i)(11 + 4i). We connect those two points with another red line. When we’re done doing this for all the red squares, a red letter “B” is complete.

43 + 80i = (8 + 3i)(8 + 7i)

76 + 65i = (8 + 3i)(11 + 4i)

68 + 87i = (8 + 5i)(11 + 4i)

58 + 103i = (8 + 5i)(11 + 6i)

46 + 125i = (8 + 7i)(11 + 6i)

We do exactly the same thing for the other colors. There are four orange squares that we factor, again drawing the lines that connect the factor pairs. An orange “O” emerges in this case.

6 + 3i = (2 + i)(3)

12 + 3i = (4 + i)(3)

10 + 11i = (4 + i)(3 + 2i)

4 + 7i = (2 + i)(3 + 2i)

Hey, look, a “T”!

47 + 140i = (8 + 7i)(12 + 7i)

79 + 100i = (10 + 7i)(10 + 3i)

Gimme an “N”!

11 + 24i = (4 + i)(4 + 5i)

19 + 34i = (6 + i)(4 + 5i)

31 + 36i = (6 + i)(6 + 5i)

And an “E.”

15 + 104i = (6 + 5i)(10 + 9i)

33 + 58i = (6 + 5i)(8 + 3i)

57 + 178i = (10 + 9i)(12 + 7i)

60 + 109i = (11 + 4i)(8 + 7i)

And, finally, another “T.”

1 + 12i = (1 + 2i)(5 + 2i)

9 + 6i = (3 + 2i)(3)

Taken in rainbow/ROYGBIV order, these colorful letters spell out the solution: BOTNET. A malicious botnet might be the reason for the “complex security” mentioned in the puzzle’s text.

Solution to last week’s Puzzle 2

Congratulations to рџ‘Џ Dan Miller рџ‘Џ of Madeira, Ohio, winner of last week’s Puzzle 2!

The second puzzle last week, by Denis Auroux, presented you with a series of “Hashiwokakuro” puzzles — a hybrid between the logic puzzle types called Hashiwokakero and Kakuro. The Earth had flooded, and only a series of small islands remained on each continent. Your job was to construct bridges connecting the islands, according to a few specific rules and the guidance of a number of signs. One sign on each continent was missing its numbers, however.

Each of the seven main continent grids could be solved uniquely with the given rules, which gave two values for the missing sums on the signs. Because we know that the answer is a word or phrase, it’s a good idea to convert those numbers into letters somehow. Indexing the missing numbers into the alphabet — 1=A, 2=B, etc. — gives two letters for each continent.

The sign’s missing values on Continent 1 are 6 and 20, or F and T.

The missing values here are 18 and 1, or R and A.

The missing values here are 16 and 18, or P and R.

The missing values here are 9 and 13, or I and M.

The missing values here are 5 and 4, or E and D.

The missing values here are 15 and 22, or O and V.

And the missing values here are 9 and 19, or I and S.

To recap, our letter pairs are FT, RA, PR, IM, ED, OV and IS. But what should we do with them now? The intercontinental bridge network will provide an all-important clue.

The final grid represented the intercontinental network, which has one twist on the previous rules: Because it represents a map of a planet, the east and west directions connect at opposite ends. (The layout of the signs ensures no passage through the north and south poles.) This is of course the only way to achieve the given sum of 10. With this piece of information in hand, the solution to the final grid is:

Notice that the numbers on the continent-islands are exactly 1, 2, 3, 4, 5, 6 and 7. Replacing each continent with the corresponding pair of letters from above, we get:

Reading in the usual order, this gives the final answer: IMPROVISED RAFT. That could come in handy in this flooded dystopia.

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  1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. Eastern time on Sunday. Have a great weekend!

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.