Puzzle Answers

Below are the answers to each of our monthly puzzles going back to January, 2025, listed in reverse chronological order. To reveal the answer to any puzzle, click the arrow below each headline. To see the original puzzle questions or crosswords, visit sciencenews.org/puzzles. Good luck!

Math Puzzle: Can you meet me at the mall?

Solution to the math puzzle from our August 2025 issue

#1. The hint suggested we consider our arrival time and our friend’s arrival time on the coordinate plane. Let x be the time you arrive at the mall (in hours after 3 p.m.), and let y be the time your friend arrives. This means x and y can take on any values between 0 and 1, with each value being equally likely.

The entire region of possible arrival times to consider is the unit square, with corners at (0, 0), (1, 0), (1, 1) and (0, 1), which has area 1. For example, (0, 0) represents you both arriving at 3 p.m., (1, 0) represents you arriving at 4 p.m. and your friend arriving at 3 p.m., etc. But not every point in the square represents you and your friend meeting each other. To successfully meet, you and your friend had to arrive within 15 minutes (or one-quarter hour) of each other. Mathematically, that meant either xy (how much earlier your friend arrived than you did) had to be at most 1/4 or yx (how much earlier you arrived than your friend did) had to be at most 1/4. These two inequalities could be written more concisely as a single inequality: |xy| ≤ 1/4.

The region that is inside the unit square (again, that was all the possible pairs of arrival times) that also satisfies the inequality we just found is the shaded hexagon below:

Another way to think about this is that, when x and y are equal, you and your friend definitely meet because you are arriving at the same time. The values of x and y are equal on the line y = x, which cuts across the diagonal of the unit square, from (0, 0) to (1, 1). You also meet your friend on any point that is within 1/4 of the line y = x in the horizontal or vertical directions. By this reasoning, we once again arrive at the hexagonal shape above.

The probability you meet your friend is the area of the hexagon divided by the area of the unit square, which is 1. You’re welcome to calculate the area of the hexagon … but a quicker way to find it is to compute the areas of triangles on either side of it (i.e., the regions where you and your friend do not meet), and then subtract the sum from 1. These triangles are right isosceles, with leg lengths of 3/4. They combine to make a square with side length 3/4, and therefore area 9/16. The area of the hexagon is 1 minus this, or 7/16. So the probability you and your friend meet is 7/16.

So, on average, what would you expect this maximum number of friends to be? The maximum is two friends with probability 7/16, and one friend with probability 9/16. Therefore, the average number of friends is 2·(7/16) + 1·(9/16) = 23/16, or 1.4375.

#2. The next problem asked you to consider three total friends. With two friends, the probability they met was the area of a hexagon inside a unit square. With three friends, you now had to consider a unit cube, with coordinates of x, y and z representing the three arrival times. All three friends meet when their arrival times are within 15 minutes of each other: |xy| ≤ 1/4, |yz| ≤ 1/4, and |zx| ≤ 1/4. These three inequalities were all true in the region below, whose volume represented the probability of all three meeting.

To find the volume of this shape, we can decompose it into a cube of edge length 1/4 in the bottom left, as well as three parallelepipeds coming off three of this smaller cube’s faces. (Similarly, the 2D case could be decomposed into a square and two parallelograms.) The cube has volume (1/4)3, or 1/64. Each parallelepiped has a square base with area (1/4)2, or 1/16, and an altitude of 3/4. So each parallelepiped has a volume of 3/64. Adding three of these to the volume of the cube gives the total volume of the figure: 10/64, or 5/32. Note that this is significantly greater than (7/16)3, since the pairwise events of friends meeting up are not independent events. (That is, if A meets up with B and B meets up with C, then it is quite likely that A also meets up with C.)

So the probability that all three friends meet up is 5/32. Next, we could try to find the probability that two friends meet up, but not three. However, that’s quite difficult to calculate!

Instead, let’s look at the probability that no friends meet each other, so that the maximum number of friends present at any given time is one. Mathematically, that’s when the following three inequalities are all true: |xy| > 1/4, |yz| > 1/4, and |zx| > 1/4. Here’s what the 3D graph of this regime looks like:

It’s a collection of six congruent tetrahedra. Each has a base that’s a right isosceles triangle with leg length 1/2, and a height of 1/2. So the volume of each is 1/48, which means their collective volume is 1/8.

If the probability that all three friends meet up is 5/32, and the probability that no friends meet up is 1/8, then the probability that a maximum of two friends meet up is what remains: 1 − 5/32 − 1/8 = 23/32.

The expected maximum number of friends is 3·(5/32) + 2·(23/32) + 1·(1/8) = 65/32, or 2.03125.It’s kind of nice that the average is more than just a pair of friends!

#3. Next up, working out the cases in a 4-D hypercube isn’t pretty. In theory, we could find the hypervolumes of the regions where one, two, three and four friends meet, and then compute the weighted average of these. But instead of doing all that, one can run thousands, or even millions, of computer simulations with four friends to pin down the expected value of the maximum that meets.

To do this, you can pick four random values between 0 and 1 and sort them. For each value x, you can check how many values among the four are within the range [x, x+0.25], which is the interval during which that friend was at the mall. For example, suppose the values are 0.2, 0.41, 0.52, and 0.79. The four intervals we’d want to check are therefore [0.2, 0.45], [0.41, 0.66], [0.52, 0.77], and [0.79, 1.04]. How many of the four values fall within each interval? Respectively, that’s 3, 2, 1 and 1. So the maximum number of friends that meet in this example is 3, which occurs sometime during the interval [0.2, 0.45].

After running 10 million such simulations, I found the average value of the maximum number of friends to be approximately 2.4766.

But to find the exact answer, we’ll need some calculus.

Suppose the arrival times of the four friends are w, x, y and z, with 0 ≤ wxyz ≤ 1. Because we’re specifying the order of the arrival times, we’re only considering a 4-D simplex within the 4-D hypercube, whose hypervolume is 1/4!, or 1/24.

Let’s look at the varying cases and work out their probabilities. First, what is the hypervolume of the region where all four friends meet? For this to happen, we need wxyzw + 0.25. The hypervolume of this region is given by a quadruple integral, which we multiply by 24 to express it as a fraction of the simplex’s volume:

Evaluating this integral gives you 13/256, so that’s the probability that all four friends meet.

Next, what is the probability that no friends meet, meaning the maximum number of friends at any given time is one? In this case, we need x > w + 0.25, y > x + 0.25 and z > y + 0.25. The volume of this region, multiplied by 24, is the following expression:

Evaluating this integral gives you 1/256. Two cases down (four friends and one friend), two to go (two friends and three friends). Between the latter two, three friends is the “friendlier” case to analyze, so let’s start there.

There are two ways three friends (but not four) can meet. First, you could have the first three friends to arrive meeting up, while the fourth arrives too late to have met the first friend. Mathematically, this meant wxyw + 0.25, along with z > w + 0.25. The probability this happens is given by:

This integral comes out to be 27/128. The other way for three friends to meet is if it’s the last three friends, while the first arrived too early to have met the third and fourth friends. (We already counted the case where the first friend meets the third.) The probability this happens is given by:

Note the slightly different order of the variables in that last integral, which comes out to 43/256. Putting these two cases together, the total probability of three friends meeting is 27/128 + 43/256 = 97/256.

At this point, we have the probability of four, one and three friends meeting up. All that’s left is two friends, the probability of which we can determine by subtracting all the other probabilities from 1: 1 − 13/256 − 1/256 − 97/256 = 145/256.

Finally, to determine the expected maximum number of friends that meet, we can use these probabilities to compute a weighted average: 4(13/256) + 3(97/256) + 2(145/256) + 1(1/256) = 317/128, or 2.4765625. This exact answer was in excellent agreement with the aforementioned simulated results.

#4. Finally, as the number of friends N gets very, very large, the average value of the maximum number that meets looks a lot like N/4. The true average will always be greater than N/4, since there will inevitably be some clumping of arrival times when they’re randomly selected. In fact, N/4 represents the minimum possible maximum number of friends that can meet — a minimum that becomes a better and better approximation for the maximum as N increases. This asymptotic result makes sense: As N becomes very large, the distribution of arrival times looks increasingly uniform across the hour, so that roughly a quarter of the total friends are present during any 15-minute window.

The “clumpiness” of arrival times — or how much greater the true average maximum is than N/4, turns out to scale with the square root of N. The graph below shows how N/4 + 0.847√N is an excellent approximation of the true behavior up to 1,000 friends. As N gets very large, the term that’s proportional to N grows much larger than the term that’s proportional to √N, so that the ratio between the average maximum number of friends who meet and N indeed approaches 1/4.

Crossword: Baby’s First Words

Solution to the crossword from our July 2025 issue
Screenshot

Math Puzzle: The conundrum of sharing

Solution to the math puzzle from our June 2025 issue

#1: How can all three friends partake in the spa using just two sheets?

The first friend lays both sheets on the mud, one on top of the other, and sweats on the upper sheet.

The second friend removes the upper sheet and uses the clean side of the lower sheet.

The third friend flips the upper sheet over (placing its sweaty side against the now-sweaty side of the lower sheet) and uses the final fresh side.

Alternatively, the third friend can discard the lower sheet and lay the sweaty side of the upper sheet directly on the mud.

#2: The next day, five friends visit the spa, and only three sheets are available. Can they all partake?

Yes! The first friend lays two sheets on the mud and sweats on the upper sheet.

The second friend removes the top sheet and lays the clean third sheet on the mud, using this one to sweat upon.

The third friend removes the top sheet and uses the clean side of the lower sheet.

The fourth friend grabs one of the two half-used sheets and places it sweaty side down. (This is called, per the brilliant and disturbing coinage of a student I met in Michigan, a “sweat sandwich.”)

The fifth friend grabs the final half-used sheet and places it sweaty side down.

#3: Soon, 10 friends visit the spa. Only five sheets are available. “Someone will have to miss out,” one of them quickly declares. “There’s no way to know that,” says another, “until we at least look for a solution.” Who’s right?

Alas, the pessimist is right. No need to look for a solution. Each of the friends needs a clean side of a sheet, plus one side must go against the mud. But that requires 11 sides, and only 10 are available. So no matter how you shuffle the sheets, someone will have to miss out.

#4: Later, the spa introduces a second kind of mud, which must not be mixed with the first. If three friends want to try both muds, how many sheets do they need at minimum? (Let’s assume each person is begrudgingly willing to lie twice on the same sheet.)

Three sheets will suffice! To begin, lay a sheet on each mud bed.

The first friend places the remaining sheet atop each mud bed in turn. (The others wait patiently.) This top sheet is then removed and set aside.

The second friend then uses the fresh sides of the two sheets already on the beds.

Finally, the third friend takes the sheet from the first friend and lays it fresh side up on each bed in turn.

#5: Lurking here is a fully general question: What’s the minimum number of sheets that allows N friends to experience M kinds of mud if each side of a sheet may touch only a single person or a single kind of mud?

It’s easiest to tackle the M = 1 case. By generalizing the answer to #2, you’ll find that (N+1)/2 sheets suffice when N is odd (with all sides of all sheets being used), and (N+2)/2 sheets suffice when N is even (with one side going unused).

The more general problem is addressed (but not fully solved) in this research paper: A. Hajnal and L. Lovász. “An algorithm to prevent the propagation of certain diseases at minimum cost,” published in Interfaces Between Computer Science and Operations Research: Proceedings of a Symposium Held at the Mathematisch Centrum Amsterdam, September 7–10, 1976.

Ben Orlin

Crossword: Limits of Knowledge

Solution to the crossword from our May 2025 issue
Solution to the May 2025 SN Crossword Puzzle

Math Puzzle: The Lesser Fool

Solution to the math puzzle from our April 2025 issue

1. Twelve thousand and twelve dollars, or eleven thousand eleven hundred and eleven dollars?

The first number is 12,000 + 12. The second number sounds smaller until your ear catches that sneaky phrase “eleven hundred”; it turns out that 11,000 + 1100 + 11 (which equals 12,111) is the larger of the two.

2. 19/200 of this pie, or 29/300 of it?

Each fraction is slightly smaller than 1/10 of the pie. The first is 1/200 shy; the second is 1/300 shy. Thus, the second fraction is closer to 1/10, and is therefore larger.

3. One kilogram of quarters, or twenty-five kilograms of pennies?

If every kilogram contained, say, 1,000 coins, then the two bundles would be equal in value. But pennies are far lighter than quarters, so you get more of them per kilogram. Thus, 25 kilograms of pennies are more valuable (albeit less convenient for shopping).

4. A penny for every second in a month, or a penny for every hour in a century?

The question can be paraphrased: To maximize ticks, would you rather your clock tick more often (every second vs. every hour) or for a longer time period (a century vs. a month)?

Since there are 60×60 seconds in an hour, the first clock ticks 3,600 times as often as the second clock.

Since there are 12×100 months in a century, the second clock ticks 1,200 times longer than the first clock.

The first value, then, is larger by a factor of roughly 3 (that is, 3600/1200), the precise value depending on which month you pick.

5. The tenth root of 10, or the cube root of 2?

These numbers are quite close together. To tell them apart, we need a way to exaggerate their differences. My proposal: Let’s raise each one to the 30th power! (This is the cube of the tenth power.)

The first number becomes the cube of 10, thus 1,000.

The second number becomes the tenth power of 2, thus 1,024.

So the second number is slightly larger.

6. $10 plus half of the second, or $20 minus half of the first?

Let’s call the value of the first envelope x.

The second envelope has 20 – 0.5x.

Meanwhile, x is equal to 10 + half of this amount.

Thus,              x = 10 + 0.5(20 – 0.5x).

This means    x = 10 + 10 – 0.25x.

Thus,               1.25x = 20.

This gives us  x = 16, and so the second envelope has $12. The first is worth more.

7. In our first year, we gained 90 percent. In our second year, we lost 50 percent. Would you rather have the amount we originally invested, or our current value?

Doubling and then losing half puts you right back where you started. But this fund didn’t quite double; instead of gaining 100 percent, it gained only 90 percent. So when it lost 50 percent, it wound up below the originally invested value. (Specifically, 5 percent below.)

Bonus: Why does the Lesser Fool always take the smaller amount? And who might be the “Greater” Fool?

The Fool takes the smaller amount because that’s how the con works! If he started taking the larger one, he’d cease to be remarkable, and that would bring an end to the parade of people offering free money. These people, naturally, are the Greater Fools.

Ben Orlin

Crossword: Buried Treasure

Solution to the crossword from our March 2025 issue
Screenshot

Math Puzzle: Imagine there’s no zero

Solution to the math puzzle from our February 2025 issue

1. What year would it be right now? For that matter, what century would it be? 

It is the year 2025, which means 2 thousands, 0 hundreds, 2 tens, and 5 ones. In the zero-less system, we cannot have zero hundreds; one of those thousands must be unboxed as ten hundreds. Thus, the current year would be 1T25. 

And what century is it? It’s the 1TXXs, which you might call “the ten-teen hundreds.” (We can’t say 1T00s, of course, because that would use the forbidden symbol.) 

Note, with perplexity, that the present century began not 25 years ago in 19TT (formerly 2000), but just 14 years ago in 1T11 (formerly 2011). And it will continue beyond the year 1T9T (formerly 2100), ending only in the year 2111 (a year that, at last, requires no translation.) 

2. Would a “six-figure salary” be more or less desirable than under the old system? 

Under our system, six-figure salaries go from $100,000 to $999,999. 

Under the new system, they go from $111,111 (the smallest six-digit number) to $TTT,TTT (the equivalent of our $1,111,110). 

So, a six-figure salary is more desirable under the new system. 

3. Map out the ways a zero-less culture would differ. Would towns commemorate 111th anniversaries? On a car’s odometer, which mileage rollover would be most exciting? And would anyone care that Wilt Chamberlain once scored 9T points in a basketball game? 

Your guesses are as good as mine, but I would wager on the following. 

  1. Just as we currently make a big deal out of birthdays and anniversaries ending in zero, we’d do the same for those ending in T. So would we celebrate a centennial on the 9Tth? No, my hunch is that the TTth (what we call 110th) or the 111th would be the big ones. 
  1. On a car’s odometer, it’s pretty clear to me that the coolest rollover is when you reach 111,111 miles, going from TT,TTT to 111,111. 
  1. People would still care that Chamberlain scored 9T in a single game, because no matter how you enumerate it, that’s impressive! 

And as for other ways a zero-less culture might differ: 

  1. We’d say that impressive sums of money have “lots of digits” (not “lots of zeros”). 
  1. The temperature that’s so cold all molecules slow to a stop would be known as “absolute freeze” or “the all-frozen” (not “absolute zero”). 
  1. The person from whom an infection spreads would be known as “the origin patient” (not “patient zero”). 
  1. “Zero-sum” games would be known as “win-lose” or “perfect tradeoff” games. 
  1. Game shows wouldn’t give away million-dollar prizes, since $999,99T is not a very cool-looking number. Instead, they’d give away $TTT,TTT or $1,111,111 prizes. 
  1. Instead of reading “0–0,” the scoreboard at the start of a game would have two blanks. 
  1. Learning arithmetic in school would be even more daunting than it already is! 

Ben Orlin

Crossword: Twisting Words

Solution to the crossword from our January 2025 issue