Welcome to the Riddler. Each week, I come up with problems related to things that are dear to us here: math, logic, and probability. Two puzzles are featured each week: the Riddler Express for those of you who want something small and the Riddler Classic for those of you who like the movement of slow puzzles. Submit a correct answer for either, and you might get a shoutout in the next column. Please wait until Monday to share your answers publicly! If you need a clue or if your favorite puzzle is gathering dust in your attic, find me on twitter.
This week, get ready to play a game of squares! You start with five shaded squares on an infinitely large grid, in the following formation:
This is generation 1. To pass from one generation to another, we consider the eight neighbors of each square (top, bottom, left, right and the four diagonal directions). Yes at least three of these squares are shaded, in the previous iteration, this square will be shaded in the next generation.
That said, here are the first four generations:
How many boxes will be shaded in Generation 10?
Additional Credit: Like NOT becomes very, very large, approximately how many squares will be shaded in the generation NOT (in terms of NOT)?
Submit your answer
From Travis Henry comes a vehicle problem:
You want to change the transmission fluid in your old van, which contains 12 liters of fluid. At the moment, the 12 pints are “old”. But changing all 12 pints at once carries the risk of transmission failure.
Instead, you decide to replace the fluid a little at a time. Every month you remove a liter of old fluid, add a liter of fresh fluid, then drive the van to mix the fluid well. (I have no idea if it’s mechanically sound, but I’ll take Travis’ word on it!) Unfortunately, after exactly one year of use, what was once fresh transmission fluid officially becomes “old.” .
You maintain this process for many, many years. One day, immediately after replacing a liter of fluid, you decide to check your transmission. What percentage of the liquid is old?
Submit your answer
Solution to last week’s Riddler Express
Congratulations to 👏 Mike Porter 👏 from Haddenham, UK, winner of last week’s Riddler Express.
Last week you had a square sheet of paper. You folded it in half along an axis parallel to two of its sides. You then folded it in half along another axis perpendicular to the original one, so that you again got a square that was a quarter of the size of the original square.
At this point, you have made three cuts along three straight lines, all the way through the folded paper. Unfolding the paper, what was the greatest number of separate pieces you could have?
After making the cuts, your folded square was made up of a number of pieces. You even knew that this number was at most seven, since it is the greatest number of regions in which the plane can be divided by three lines. But what needed to be determined was how many times each of these seven regions could be duplicated by folding the paper before cutting it.
Consider the following example, in which the paper has been folded in the upper right quarter, after which you have made the three cuts shown:
When you unfold the paper, there would be a copy of the region labeled “1”. Meanwhile, the regions labeled “2” would each have two copies, horizontal or vertical reflections of each other. Finally, the region labeled “4” would have four copies, one in each of the four quarters of the larger square.
The key to this puzzle was realizing that these copy numbers (one, two, and four) were determined by the sides of the smaller square—specifically the left and bottom sides—delimiting which region. If a region was bounded by both the left and bottom sides, there was only one copy. If it was bounded by exactly one of the two sides, there were two copies. And if it was not bounded by any side, there were four copies.
The riddle asked you to maximize the number of copies, which meant you wanted a lot of regions with four copies and not so many regions with one copy. Now at least one region had to contain the lower left corner of the smallest square, which implied that it would result in a copy. But what about the other six regions? They could all have four copies, if you made the cuts as follows:
This gave you a maximum of 25 copies.
For extra credit, instead of three cuts, you did NOT cuts. Now what was the highest number of coins you could have?
Mike, this week’s winner, extended the approach to three cuts, again having only one region bounded by the inner sides of the smaller square:
In fgeneral, NOT lines can divide the plane into at most NOT(NOT+1)/2+1 regions. One of these regions should have a copy, while the others NOT(NOT+1)/2 had four. The maximum total was therefore 4NOT(NOT+1)/2+1, or 2NOT(NOT+1)+1 copies.
By the way, this quadratic relationship means you can generate coin loads without doing too much many cuts. For example if you have the stamina and precision required to make 100 cuts, you can generate up to 20,201 coins!
Solution to last week’s Riddler Classic
Congratulations to 👏 Steve Curry 👏 from Albuquerque, New Mexico, winner of last week’s Riddler Classic.
Last week, The Riddler readers Ellora Sarkar and Daniel Gomez got married! In keeping with the hexagonal design of their wedding backdrop, I presented the following puzzle:
The largest regular hexagon in the diagram below had a side length of 1.
What was the side length of the smallest regular hexagon?
Solver David Ding assumed that half the side length of the smaller hexagon was X. Next, David drew the following right triangle inside the circle:
The hypotenuse of this triangle was the radius of the circle, which also happened to be the length of the side of the larger hexagon, 1. The shorter leg of the triangle, which was the left half of the taller side of the smaller hexagon , was X/2. Meanwhile, the length of the remaining leg, which could be found with a bit of trigonometry, was √3/2 + √3Xor √3(X + 1/2). Then David plugged these three values into the Pythagorean theorem, which gave the following quadratic equation (after some simplifications): 13X2 + 12X − 1 = 0. This could be factored as (13X−1)(X+1) = 0. Rejecting the negative root meant that Xthe length of the side of the smaller hexagon, was 1/13. It was pretty neat to see that after all that work, a nice little rational number appeared!
For more credit, you were asked to find the side lengths of two other, even smaller, hexagons on top, as shown in the animation below:
Applying the Pythagorean theorem again, but this time to smaller and smaller hexagons, solver Laurent Lessard found a recurrence relation for the side length of a hexagon (Xnot+1) as a function of the side length of the largest adjacent hexagon (Xnot):
Xnot+1 = (√(48+Xnot2) − √(48−12Xnot2))/13
In the puzzle you were told that X1 was 1. And we just found out that X2 was 1/13. By applying the above formula, X3 was approximately 4.27×10-4 and X4 was approximately 1.32×10-8.
You may have noticed that these values have become very small, very quickly. As Laurent showed, the number of hexagons and the length of the sides of the smaller hexagon approximated a double exponential relationship.
This is the kind of detail you need for a successful wedding. Congratulations again to Ellora and Daniel!
Want more puzzles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles in this column and some never-before-seen puzzles. It’s called “The Riddler”, and it’s in stores now!
Want to submit a puzzle?
Email Zach Wissner-Gross at [email protected]