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.
From Diarmuid Early, a daily dose puzzle:
You and your spouse each take two gummy vitamins every day. You share a single bottle of 60 vitamins, which come in two flavors. You each prefer a different flavor, but it seems childish to fish out two of each type (but not to take gummy vitamins). So you just take the first four that fall out and then divide them up according to your preferences. For example, if there are two of each flavor, you and your spouse get the vitamins you prefer, but if three of your preferred flavor come out, you get two of the ones you like and your spouse will get one of each.
The question is, on average, what percentage of the vitamins you take are the flavor you prefer? (Assume that the bottle starts out with a 50-50 split between flavors, and that the four selected each day are selected uniformly at random.)
From Dan Schauer, inspired by a recent physics paper and a blog post by Sabine Hossenfelder, a difficult problem of insects and landscaping:
A grasshopper lands somewhere randomly on your lawn, which has an area of 1 square meter. As soon as it lands, it jumps 30 centimeters. What shape should your lawn be to maximize the chances that the grasshopper will still be on the lawn after the 30-centimeter jump? (Hint: It’s not a circle.) This is a visual problem, so submitting a link to your shape may be best, although descriptions are also fine.
Extra credit: What if the grasshopper jumps X centimeters instead?
Solution to last week’s Riddler Express
Congratulations to 👏 Russell Wang 👏 of New York, winner of last week’s Riddler Express!
Last week, the Senate voted 81-18 to end the government shutdown. The vote tally caught the Riddler’s attention because it’s a palindrome, meaning that it reads the same forward and backward. That specific palindrome was made possible by the absence of one senator, John McCain. Your challenge last week was to identify whether absences were necessary to create palindromic votes in the Senate and, if so, what numbers of absences did the trick.
The answer depends on whether we count the hyphen as part of the tally. If we ignore it, senators need not be absent to create a palindromic vote. The tallies 91-9 and 5-95 are examples with all 100 senators voting. If we count the hyphen, as was my intention, we’re more constrained. (In this case, “91-9” is “9-19” in reverse — which is different from the original.) Here, a palindromic vote tally takes the form AB-BA, where A and B represent single-digit natural numbers. There’s a way to turn that into a math problem: the total number of votes is 10A+B+10B+A, or, simplified: 11(A+B). The number of senators voting, therefore, has to be a multiple of 11.2 Put another way: The number of senators voting has to itself be a palindrome — 99, 88, 77 and so on.
Here is a list of the possible palindromic vote tallies, given for the different counts of voting senators:
99: 81-18, 72-27, 63-36, 54-45, 45-54, 36-63, 27-72, 18-81
88: 71-17, 62-26, 53-35, 44-44, 35-53, 26-62, 17-71
77: 61-16, 52-25, 43-34, 34-43, 25-52, 16-61
66: 51-15, 42-24, 33-33, 24-42, 15-51
55: 41-14, 32-23, 23-32, 14-41
44: 31-13, 22-22, 13-31
33: 21-12, 12-21
I suppose you could also include those bizarre cases in which only a single-digit number of senators took each side of a vote: 1-1, 2-2, 3-3, 4-4, 5-5, 6-6, 7-7, 8-8 and 9-9. And if we ignore the hyphen, we can add these tallies to the list above: 90-9, 80-8, 70-7, etc.
For extra credit, a few solvers found 247 examples of palindromic votes in the past 30 years of Senate history. Another solver just told me to “ask Harry Enten.” Excuse me a moment, Riddler Nation, I think I see him prowling the newsroom. Hey, Harry …
Solution to last week’s Riddler Classic
Congratulations to 👏 Marc and Sam Surette 👏 of Fairfax, Virginia, winners of last week’s Riddler Classic!
Last week’s problem was a mashup of two famous mathematical ideas, separated by millennia: Archimedes’s Ostomachion puzzle from ancient Greece and the four-color theorem, first proven in the 20th century. Given the Ostomachion square shown below, you were asked to color it with four colors such that no bordering regions shared a color and that each color shaded a region of equal area — 36 square units in this case. For extra credit, I asked how many different ways there were to complete this challenge.
Finding one such coloring isn’t too difficult and could be accomplished through trial and error. It’s straightforward to calculate the areas of the Ostomachion pieces using the formula for the area of a triangle: A = ½ bh. The areas all turn out to be tidy whole numbers: 3, 6, 9, 12, 21 or 24.
From there, you need to find coloring combinations that satisfy two constraints: each color’s total area adds up to 36 squares and all neighboring colors are different. Here’s one example, submitted by Tenzin Larkin, age 8:
And here’s a totally different example, submitted by Guy Moore, age 48:
Determining exactly how many different arrangements could do the job was more complicated, and here, most solvers turned to computer analyses. There are nine distinct solutions. Keith Hudson found them using some Perl code and animated them in the lovely pastel gif below:
As it turns out, you could have solved this challenge with only three colors. There were two ways to do that, as illustrated by Laurent Lessard:
Want to submit a riddle?
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