Problem Circle Solution
Solving an easier problem is a huge key to making progress with the Frogs and Toads problem from the Summer 2013 MTCircular. With the 5 by 5 board, there are a lot of moves in the solution, and lots of places where one false move leaves you stuck many moves later without any easy way to see just where you went wrong. However, if you simplify things to 1 by 5 or even 1 by 3, then you can solve those problems easily. Solving an easier problem is always a good strategy for getting started on anything, but as we’ll see, in this problem it is even more powerful than usual. Read more…
Solutions to Past Problem Circles
In the Problem Circle from the Winter 2013 issue of MTCircular, we asked for integer solutions to the equation xxyy = zz. One such solution is x = 21236, y = 2838, and z = 21137. Our offer of a free book still stands for the best additional solution sent to problemcircle (at) aimath.org.
In the Problem Circle from the Autumn 2012 issue of MTCircular, Joshua Zucker posed the following problem:
Start with a list of the two numbers: 1, 1. Then, from that list, generate a new list by inserting the sum of each pair of numbers between those two numbers. Continue to generate new lists in this way. Thus, the next lists are 1, 2, 1 and 1, 3, 2, 3, 1 and so on. How many 2012s are found in the 2012th list?
Page 17 of the Winter 2013 issue of MTCircular includes a description of work by Mary Fay-Zenk and Richard Grassl on this problem.
Joshua Zucker writes, “My favorite introduction to the magic of this sequence comes from Brent Yorgey’s Math Less Traveled blog, which has a series of about a dozen posts on two of the most interesting aspects of this sequence. First, if you leave off the 1 at the end of each row, the nth number you encounter is the number of ways of writing n as the sum of at most 2 copies of each power of 2. For instance, the tenth number is 5 because 10 = 8 + 2 = 8 + 1 + 1 = 4 + 4 + 2 = 4 + 4 + 1 + 1 = 4 + 2 + 2 + 1 + 1 gives 5 such representations. Secondly, if you take the ratios of successive terms, you find that consecutive terms are relatively prime, and what’s more, every positive rational appears exactly once!
“Because of this result, the number of columns of 2012s is given by the number of ways of writing 2012 as the sum of two relatively prime numbers, which is also equal to the number of simplified fractions with denominator 2012.
“For more, take a look at the links in the Online Encyclopedia of Integer Sequences. In particular, the amazing fraction result comes from a famous paper of Calkin and Wilf that is quite readable for nonmathematicians though it was written for a mathematical audience. There is also plenty of further reading (e.g., http://oeis.org/wiki/User:Peter_Luschny/SternsDiatomic and http://www.cs.utexas.edu/users/EWD/ewd05xx/EWD578.PDF) that could be done on this topic!”
In the Problem Circle from the Winter 2012 issue of MTCircular, we asked you to explore counting, adding, multiplying, and writing decimals in base -4. Recall that in base -4, the place values from right to left are the powers of -4: 1, -4, 16, -64, 256, etc. Also, each number should be represented using only the digits 1, 2, 3, and 0 (which is equivalent to 4), with no negative signs.
Joshua Zucker has helped get us started on this problem by illustrating how to multiply 7 and 9 in base -4.
First, note that
Then you can set up the problem:
To check our answer:
The trickiest part of this multiplication is the carrying. Specifically, why have we carried -1s? The key is to remember that we are actually counting groups of -4. Every time we get back to 0 (4), we have actually completed counting -1 groups of -4. For example, to determine the tens place in the problem above, we needed to add 3 and 2. To represent 5, we wrote down a 1 in the tens place and carried a -1. This makes sense if you think of the 5 as 1 plus -1 groups of -4, or 1 + (-1 x -4). Combining this understanding of how to carry with the notion of borrowing will help you to see how we got to the representations of 7 and 9 in the first place.
In the Problem Circle from the Summer 2011 issue of MTCircular, we asked whether it is possible to paint each point in the plane, using three different colors, so that every straight line in the plane has exactly two different colors. Here are two nicely different solutions to this problem, courtesy of Hema Gopalakrishnan (top) and Joshua Zucker (bottom).
For her solution at the top, Hema Gopalakrishnan writes, “Using the XY-coordinate system to represent points in the plane, suppose the origin is colored green, points on the X-axis other than the origin are colored red and points on the Y-axis other than the origin are colored blue. Further suppose that the points in the 1st and 3rd quadrants are colored blue and points in the 2nd and 4th quadrant are colored red. Note that the only point colored green is the origin. Then every line through the origin will have two colors, blue and green or red and green. Every line not passing through the origin will have the two colors blue and red.”
For his solution at the bottom, Joshua Zucker writes, “Any line not passing through the intersection point can’t be parallel to both of the intersecting lines, so it will be green and red. Any line passing through the intersection point either is one of the intersecting lines, in which case it is red and blue, or meets the intersecting lines at only the intersection point, so it is green and blue.”