Table of Contents:
-- Mathematical Mystery Tales
-- Putting Your Logic to Work
-- The Games Mathematicians Play
-- Some Mathematical Recreations
-- Magic Squares & More Recreations
-- The Konigsberg Bridge Problem Update
-- Checkerboard Mania
-- A Few Oldies
This chapter sounds like fun! But unfortunately, we've reached the weekend, and so we have to wait until next week to play any of these games. Instead, let's celebrate the fall equinox today by continuing in Ogilvy's book.
Chapter 7 of Stanley Ogilvy's Excursions in Number Theory is called "Number Curios." It begins:
"It is assuredly curious that 142857, when multiplied respectively by 2, 3, 4, 5, and 6 yields always a product consisting of a cyclic permutation of the original six digits. But we analyzed this behavior in Chapter 5, answered the question 'why,' and examined the problem in connection with other allied problems."
Ogilvy tells us that a number "curio" is a "curiosity" -- something interesting that happens with numbers as a mere coincidence. He explains that the property of 142857 is not a coincidence, and he explained why in an earlier chapter (i.e., it's the period of 1/7). Instead, his first true curio is:
a^b * c^d = abcd (read as four digits, not multiplication)
The lone solution was found by mathematician and author Joseph de Grazia:
2^5 * 9^2 = 2592
which he finds by trial and error. Ogilvy tells us that some curios are base-dependent -- so the above example depends on the decimal system. It's obvious, for example, that a^b * c^d = abcd has no solution in binary -- the only digits are 0 and 1, and the only powers of 0 and 1 are 0 and 1, so we can never get a four-digit answer.
Ogilvy gives us the following fraction:
16/64 = 1/4
where the 6's "cancel," and asks whether there are any more examples aside from the trivial:
44/44 = 4/4 = 1
where the 4's "cancel." To solve this, Ogilvy lets the two-digit numbers be xy and yz, where y is the digit that "cancels." We write this "cancellation" formally as:
(10x + y)/(10y + z) = x/z,
which says that
9xz = y(10x - z).
Now it's possible that the 9 on the left-hand side divides the (10x - z) on the right-hand side. Then:
10x - z == 0 (mod 9)
10x == z (mod 9)
But then
10 == 1 (mod 9)
10x == x (mod 9)
which implies that z == x (since both are nonzero digits), and this is trivial (like 44/44). So the only interesting case is if 9 doesn't divide (10x - z). Since 9 isn't prime, this doesn't imply that 9 must divide the other factor y on the right-hand side, but it does imply that 9 and y have a factor in common, such as 3. We then use trial-and-error on all cases, since only 3's, 6's, and 9's can cancel:
26/65, 19/95, and 49/98
The author also gives us an example where two digits cancel:
143185/1701856 = 1435/17056.
Even though Ogilvy doesn't give examples in other bases, I want to consider them now. Yes, I know it's amazing that I keep returning to alternate number bases over and over in this book. Over the summer, I spend a day or two on number bases in reading the Pappas book. It's one of my favorite topics, yet I didn't do it justice over the summer, so I'm squeezing it in during Ogilvy's book.
The equation from earlier:
9xz = y(10x - z)
generalizes to other number bases as:
omega * xz = y(base * x - z)
Except for trivial solutions, y and omega (that is, base - 1) have a factor other than 1 in common. In fact, I notice that in the case where y = omega, the decimal examples 19/95 and 49/98 generalize to many bases. For each divisor of the base (other than 1 and the base itself), let x be one less than the divisor and z be the divisor subtracted from the base. This works in all bases other than prime bases, since prime bases have no divisors other than 1 and the base itself:
Quaternary: 13/32 = 1/2
Senary: 15/53 = 1/3, 25/54 = 2/4
Octal: 17/74 = 1/4, 37/76 = 3/6
Dozenal: 1b/b6 = 1/6, 2b/b8 = 2/8, 3b/b9 = 3/9, 5b/ba = 5/a
14-mal: 1d/d7 = 1/7, 6d/dc = 6/c
All of these bases have prime omegas, and so no others are possible in these bases.
For bases with composite omegas, there are other possibilities for y, due to the requirement that the y and omega not be relatively prime. In quinary, y = 2 is possible, but a quick check of all possible numbers shows us that 2 doesn't work. In septimal, y = 2, 3, 4 are all possible, but as it turns out, none of these work either. In nonary, something interesting happens. Since the base itself is composite, it has an example where the omega 8 cancels, 28/86 = 2/6. But this can be reduced to 14/43 = 1/3:
Nonary: 14/43 = 1/3, 28/86 = 2/6
I ran a short TI program to find more examples. The next odd base with cancellations is 15-mal (pentadecimal), which figures since both the base and its omega are composite. It's the first base to have six cancelling fractions:
Pentadecimal: 17/75, 27/76, 2c/c9, 2e/e9, 3c/ca, 4e/ec
And hexadecimal (used in computers) beats base 15. It's the first base with seven fractions:
Hexadecimal: 15/54, 19/96, 1f/f8, 2a/a8, 39/98, 3f/fc, 7f/fe
There's no need to go any further, since by this point we're beyond the human scale. I do point out that I was unable to find any prime base with a cancelling fraction, even though theoretically they should exist since they have composite omegas.
Returning to Ogilvy, his next curio involves perfect numbers:
6, 28, 496, 8128, 33550336, 8589869056, ...
As it turns out, when we repeatedly add the digits of any perfect number, the answer is always 1 -- except for the first perfect number 6. Ogilvy asks, why is the digital root always 1, and why is 6 the only exception?
To answer this, we recall that the perfect numbers are of the form 2^(p-1)(2^p - 1), p prime. We begin by working in mod 3, noting that if p is an odd prime then p - 1 is even:
2 == -1 (mod 3)
Therefore 2^(p-1) == 1 (mod 3)
So 2^(p - 1) = 3k + 1 for some k. Then we multiply by 2:
2^p = 6k + 2
2^p - 1 = 6k + 1
2^(p-1)(2^p - 1) = (3k + 1)(6k + 1)
= 18k^2 + 9k + 1 == (1 mod 9)
2^(p-1)(2^p - 1) - 1 == 0 (mod 9)
Since we're thinking about mod 9, we consider the omega rule for 9. Multiples of 9 have a digital root of 9, and so numbers with a digital root of 1 are == 1 mod 9. So we just proved that perfect numbers have a digital root of 1.
That is, except for the perfect number 6. The reason for the exception is that we assumed that p is an odd prime, while the perfect number 6 corresponds to the prime p = 2, which is not an odd prime.
In the next example, Ogilvy notes that the product of a two-digit number and its reversal (57 * 75) is never a square, except in the trivial case where a number is its own reversal (55). He tells us that there are non-trivial examples of square products of longer numbers and their reversals:
169 * 961 = 162409 = 403^2
1089 * 9801 = 10673289 = 3267^2
But in case, the number and its reversal are both squares themselves. Ogilvy conjectures that the product of a number and its reversal is never a square unless both factors are squares or palindromes.
The next example is definitely base-dependent. Our base 10, factors as 2 * 5, and so its possible to factor the first few powers of 10 without any zeros in the factors:
10^2 = 2^2 * 5^2 = 4 * 25
10^3 = 2^3 * 5^3 = 8 * 125
Ogilvy writes that the only other known powers of 10 that factor without zeros are 4, 5, 6, 7, 9, 18, and finally 33. But 10^8 doesn't work because 5^8 = 390625 has a zero. As the powers gain more digits, it's more likely that a zero will show up in either 2^n or 5^n, and so it's likely that there are no more powers of 10 that can factor without zeros.
The author continues looking at the digits of 2^n and 5^n:
2, 4, 8, 16, 32, 64, 128, 256, 512, ...
He notices that the final digits are periodic: 2-4-8-6. Moreover, the tens, hundreds, and further digits are also periodic, but they take longer to repeat. Ogilvy tells us that the digits in the nth place from the right repeat with period 4*5^(n - 1).
5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125, ...
Obviously all the units digits are 5, and all the tens digits are 2. Technically, this means that both are periodic with period 1. The hundreds digits are seen to repeat 1-6 (period 2), and as it turns out, the thousands digits repeat 3-5-8-0 (period 4). The general period is 1/2*2^(n - 1) for n > 1. That is,
5^(n - 1) controls the period length in the powers of 2
2^(n - 1) controls the period length in the powers of 5
Obviously, in other bases besides 10 = 2 * 5, the powers of the factors of the base follow somewhat different patterns.
In the next example (which is base-independent), Ogilvy tries to find integers to solve:
(a^2 + b^2)(c^2 + d^2) = 2
The simplest solution is:
(2^2 + 4^2)/(1^2 + 3^2) = 2
The author gives us a more spectacular curio:
(3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2)/(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2) = 2
In the final example of the chapter, Ogilvy provides some more patterns:
1 + 2 = 3
4 + 5 + 6 = 7 + 8
9 + 10 + 11 + 12 = 13 + 14 + 15
16 + 17 + 18 + 19 + 20 = 21 + 22 + 23 + 24
There's a similar pattern for sums of squares:
3^2 + 4^2 = 5^2
10^2 + 11^2 + 12^2 = 13^2 + 14^2
21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2
36^2 + 37^2 + 38^2 + 39^2 + 40^2 = 41^2 + 42^2 + 43^2 + 44^2
The author tells us that in the first sequence, the last term before the equal sign in the n'th row is given as n(n + 1), and in the second, the last term before equals is [2n(n + 1)]^2. So then we wonder whether there is a similar pattern for cubes with [3n(n + 1)]^2. But then he points out that the first equation of that sequence would be:
5^3 + 6^3 = 7^3
which contradicts Fermat's Last Theorem.
Ogilvy uses this as an excuse to write about FLT a little more. Even though the equation above is false, the two sides really are close:
5^3 + 6^3 = 341
7^3 = 343
He asks, why is it that:
x^n + y^n = z^n
has infinitely many solutions for n = 1 and n = 2, yet no solutions for higher n? His answer is:
"This is the question that has baffled mathematicians for centuries."
That should be had baffled, of course, until Andrew Wiles came along.
The case n = 2 is associated with triangles -- right triangles, to be exact. In today's lesson, students learn about triangles and other polygons.
Lesson 2-7 of the U of Chicago text is called "Terms Associated with Polygons." (It appears as Lesson 2-6 in the modern edition of the text.)
This is what I wrote two years ago about today's lesson:
Lesson 2-7 of the U of Chicago text deals with polygons. Notice that this lesson consists almost entirely of definitions and examples. But this chapter was setting up for this lesson, since a polygon is defined (Lesson 2-5, Definitions) in terms of unions (Lesson 2-6, Unions and Intersections) of segments:
A polygon is the union of three or more segments in the same plane such that each segment intersects exactly two others, one at each of its endpoints.
It follows that this section will be very tough on -- but very important for -- English learners. I made sure that there is plenty of room for the students to include both examples and non-examples of polygons. The names of n-gons for various values of n -- given as a list in the text -- will be given in a chart on my worksheet.
The text moves on to define a polygonal region. Many people -- students and teachers alike -- often abuse the term polygon by using it to refer to both the polygon and the polygonal region (which contains both the polygon and its interior). Indeed, even this book does it -- when we reach the chapter on area. Technically, triangles don't have areas -- triangular regions have areas -- but nearly every textbook refers to the "area of a triangle," not the "area of a triangular region." Our text mentions polygonal regions to define the convexity of a polygon -- in particular, if the polygonal region is convex (that is, if any segment whose endpoints lie in the region lies entirely in the region), then the polygon itself is convex.
The text then proceeds to define equilateral, isosceles, and scalene triangles. A triangle hierarchy is shown -- probably to prepare students for the more complicated quadrilateral hierarchy in a later chapter.
Many math teachers who write blogs say that they sometimes show YouTube videos in class. Here is one that gives a song about the three types of triangle. It comes from a TV show from my youth -- a PBS show called "Square One TV." This show contains several songs that may be appropriate for various levels of math, but I don't believe that I've ever seen any teacher recommend them for the classroom. I suspect it's because a teacher has to be exactly the correct age to have been in the target demographic when the show first aired and therefore have fond memories of the show. So let me be the first to recommend this link:
Another song from Square One TV that's relevant to this lesson is "Shape Up." Notice that many geometric figures appear on the singer's head -- though not every shape appearing on her head is a polygon:
2017 Update: Today is an activity day. Two years ago, right after Lesson 2-7, I posted activities for the Daffynition Game and Jeopardy, and so I repeat those activities today. This is what I wrote two years ago about these activities:
And now I present my worksheet for the Daffynition Game. Remember that only one of these worksheets need to be given to each group -- in particular, to the scorekeeper in each group. The students write their guesses for Rounds 1-4 (or 5) on their own separate sheet of paper. I recommend that it be torn into strips so that they are harder to recognize. And the teacher provides the index cards, one for each student. Make sure that the students give back the index cards so you can reuse them for the next period. The students may keep their "guess cards," so there should be one for every student in every period.
The second page is for the Jeopardy game -- just as with the Daffynition game, there should be index cards, with the number of points on one side and the question (um, the answer, since the response is the question) on the other. In my class the questions were taped to the front board. In this version of the game, the categories correspond to the four lessons covered earlier this week. Of course, some lessons, such as Lessons 3-1 and 3-2 on angles, are tailor-made for Jeopardy, but unfortunately we haven't quite covered the lesson. Of course, we'll get there next week. I didn't include a Final Jeopardy Question, but here's a tricky one:
Final Jeopardy Category: Types of Polygons
If two points lie in the interior of this type of polygon, then the segment joining them lies in the interior.
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