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Module 5 – Chapter 7: Mathematical Curios and Chapter 8: Primes as Leftover Scrap
Lots of different topics – most are unusual and fun!
Here’s a tiny taste:
Look at the number 2, 592. Did you know that it can be written in factored form as ?
Let’s explore how many other numbers can be written this way! The first step is to come up with an equation that expresses the relationship – since there are 4 different digits, I’ll use a, b, c, and d.
a, bcd = expresses what we’re trying to find.
Now let’s see how many numbers there are like this. We’ll use counting methods
___, ______. There are 6, 561 digits to sort through.
Clearly we need to attack this abstractly and not from a list! We’ll leave that effort for another time!
Now let’s look at a problem that vexes math teachers who are teaching division in fractions – especially that you may cancel common factors, but not anything else.
“Wrongly cancelling”
If you have
And you spend time telling them how to cancel common factors…
Then one of them may show up with this example:
\
And you’re forced to admit that cancelling the 6’s “works” because
And there are a few others that work like this, too. But it’s a fluke, not a consistent rule.
If you try it with . You can point out that this type of cancelling doesn’t ALWAYS work. Let’s explore the situations in which it DOES work.
First we’ll set up our equation:
Now it’s NOT x times y or y times z in the numerator and denominator!
It’s and x, y, and z are single digits! And the y “cancels wrongly”.
Popper 09, Question 1
Here’s another one that works with the wrong cancelling:
A True
B False
It’s and x, y, and z are single digits! And the y “cancels wrongly”.
Here’s the plan of attack:
We’ll look at 3 cases:
Case 1 x = y = z
Case 2
Case 2A Suppose 9 divides 10x – y
Case 2B Suppose 9 divides y
Case 2C Suppose 3 divides y
And we’ll have to do some exhaustive reasoning because there are not any theorems about this to help us along.
Now let’s look at Case 1 x = y = z
This one works but isn’t very interesting. If all the digits are the SAME, then you get fractions that look like:
Well, yeah: any number divided by itself!
However Case 2 offers some interesting work!
Case 2
Let’s look at our fraction again and do some algebra:
Because I’m in Case 2, I know that each variable is a SINGLE digit.
Now note that both sides of our new equation are products – this means that I can do some dividing. 9 can go into either of the two factors on the right hand side.
Now for our subcases:
A Suppose 9 divides 10x – y.
This means which means that 10x – z is a multiple of 9
Now:
Add z to both sides:
Note that so we’ll replace 10 with 1 in the equivalence:
And we have a contradiction: Case 2 is with x, y, and z single digits…thus we find here that x = y since they’re single digits and in the SAME equivalence class .
Case 2A is a “no go” and 9 does not divide 10x – z
B. Suppose 9 divides y
Then 9 = y because these are all single digits!
Now since 9 = y, we know that neither x nor z is equal to 9.
2B cases
Suppose x = 1, substitute into the equation:
x = 1, y = 9, z = 5 works!
Suppose x = 2, substitute into the equation:
Nothing
Suppose x = 3
4z = 30
Suppose x = 4
Perfect! Another 1:
It turns out that no other x values produce anything…I checked so let’s skip to the last case:
2C Suppose it’s not 9, but 3 that divides y
So y = 3 or y = 6
The first one, doing trial and error with different x’s yields nothing, but with y = 6
If x = 1 we get
If x = 2 we get
So there are only 4 of these that work “wrongly” in the whole universe of double digit fractions.
But there are more versions of this kind of thing if you raise the number of digits you allow:
!
The reasons for showing this are several:
I want you to see a bit more of how math research and exploration go in a guided situation AND I wanted us to explore the structure of the problem so you can see an illustration of math reasoning at work.
On to Digital Sums and Perfect Numbers.
A digital sum is the sum of the digits taken down to a single digit.
For example:
When you find the digital sum of a perfect number greater than 6, you always get 1.
Now NOT every number with a digital sum of 1 is perfect, while all perfect numbers
do sum to 1. Why is this true?
We’ll explore this after
Popper 9, Question 2
warning: this one is vicious and cruel…be sure to go far enough!
The digital sum of any repunit, , is n the index of the repunit.
A true
B false
Now back to some earlier pages:
Page 13: If a number is divisible by 9, the sum of the digits is divisible by 9.
i.e. the digital sum
Page 22: A perfect number equals the sum of its proper divisors plus 1.
Page 22: An even number N is perfect whenever for P a prime number.
Fact: All primes greater than 2 are ODD numbers.
Fact: is always an even number and we can show the following
while recall [2]=[−1]
Now for
Now if P is an odd prime: *
So now let’s multiply both sides by 2:
Let’s look back a N, our perfect number:
and substitute in some algebra formulas for the factors: * and **
AHA! Now we have it. We have 9 digits with 9 equivalence classes and N is in equivalence class 1.
Now, one other thing to notice is that digital sums put each number in one of 9 equivalence classes [0] to [8] mod 9. So digital sums are yet another partition on the natural numbers.
Amazing! And all the perfect numbers are in [1] mod 9.
Popper 09, Question 3
The digital sum of 287 is 8.
A True
B False
Two digit reversal products page 88
It has been shown that if you have a double digit number ab where a is not equal to b, then the product of ab and ba is NEVER a perfect square.
For example 15(51) = 765, which is not a perfect square.
Now if you’re being careless you’ll say ab(ba) = abba = because multiplication commutes.
BUT ab = 10a + b and ba = 10b + a…and you have to use FOIL on them and you DON’T get abba like in a times b times b times a… you get .
Now this is not a true statement for larger digit numbers:
169(961)=162409 =
The key here is that each of the number and it’s reversal are themselves perfect squares so some rearranging can be done
169 = 13(13)
961 = 31(31)
13(13)(31)(31) = 403(403) with 13(31) = 403
This is a conjecture and not a proof, by the way.
Zero-free factors of powers of 10 – page 89
Periodicity with successive powers of 2
Note the cycle of the end digits:
2, 4, 8, 16
32, 64, 128, 256
512, …
This cycle is easy to see. The 10’s digits have a cycle that is 20 long. The hundreds digits have a cycle that is . In fact, every digit place has a cycle that is long where n is the number of places to the right that the digit is. n = 2 for the 10’s digit.
Now powers of 5 have exactly the SAME situation for cycles of digits except that the length of the cycle is computed with with n being the digits place from the right.
The reciprocity of 2 and 5 in the calculation is an artifact of our base 10 digit system and due to the fact that 10 has 2 proper factors: 2 and 5.
Popper 09, Question 4
The 100’s digit in the powers of 5 have a repeat period of
A. 2
B. 4
C. 6
Perfect square sums in fractions that are equal to 2: page 90
How many solutions are there? LOTS!
Here’s one:
Here’s one not in the book:
In the homework you have to find some of these and come up with some tips for finding them.
Adding patterns Page 91
Let’s take row 2 and see why this works
6 = 2(3)
And look at row 3
12 = 3(4)
Popper 9, Question 5
The next row is:
A There is no next row – the pattern stops there
B
Second video starts here.
Chapter 8 pages 93 – 109
We start off with Primes – this time with an eye toward how they are sprinkled in the number line. We’ll be doing “gap analysis” – studying the spaces between the primes.
Starting with 2, then 3, then 2 spaces to 5 and 2 spaces to 7 then 3 spaces to 11….
Primes get further and further apart as they get bigger and bigger.
We have estimators of how many primes there are less than a given x. These are, of course, educated guesses. We also have ways for finding spaces of a certain size between two primes.
First let’s look at the estimators:
It’s easy to just count when the numbers are small. There are 4 primes smaller than 10
{2, 3, 5, 7} but how many are there less than ? We don’t like the idea of counting for this, we want to estimate.
One reasonable estimator is: where x is the boundary number and “ln x” is a function you can find on any scientific calculator – “ln(x)” is the “natural logarithm of x”.
Suppose we want the number of primes less than 9
an irrational number that we’ll round to 4
{2, 3, 5, 7} pretty good. Now sometimes it’s high or sometimes it’s low.
If x = 100 the estimator gives us 22 and in fact it’s 26 primes…so it’s a little low.
A second estimator is by Dr. Liouville (the transcendental number guy!). It involves math way beyond what we’ll be doing:
This is an integral and it finds the area under a curve!
Popper 10, Question 1
Given the estimator formula: and the fact that ln(12) 2.5, we can say that the number of primes less than 12 is approximately:
A 4
B 5
C 6
Now on to “gap analysis”. Suppose we want at least 5 composite numbers between two primes:
Prime1 ______… Prime2 “at least 5” is 5 or 6 or 7…”
How can I guarantee this? Well there is a handy algorithm for doing this (an algorithm is a series of steps that use math to achieve an desired outcome).
Let’s look at (n + 1). This is a number
These are ALL composite numbers.
Example: Suppose I use n = 5 … n+1 = 6
This is a composite number
Here are 5 composite numbers and they’re consecutive!
So if I find the closest prim to 6! +1 and the one closest to but larger than 6!+6…I’ve got at least 5 composite numbers between these.
719 is the largest and 727 is the smallest prime greater than 6!+6.
The gap algorithm doesn’t guarantee exactly n composites inbetween primes…it guarantees at least n. Neither does it guarantee that you get the very first gag of size n, either.
The book’s example is a gap of at least 100…a very big number indeed.
Another example: a gap of at least 3 composites
n = 3 and we’ll use n + 1 = 4
4!+2 = 26
4!+3 = 27
4!+4 = 28
23 is the greatest on the low end and 29 is the smallest on the high end.
Popper 10, Question 2
To find a gap of at least 75, start with 75!+1
A True
B False
Let’s look at regularly spaced numbers – specifically arithmetic sequences:
With an arithmetic sequence you have an initial number, a, and a difference d. Each term of the sequence is “d” away from the next term.
Suppose a = 5 and d = 2
5, 7, 9, 11, 13, 15, …and so on.
Is it possible to make an arithmetic sequence with NO primes in the list?
Yes! It’s actually quite easy. The key is that the greatest common divisor of a and d is NOT 1.
Which is to say that they are NOT relatively prime.
Note in the example above: (5, 2) = 1 and there’s 7, 11, 17…in the list.
However if we take a = 10 and d = 5 ( (10, 5) = 5)
10, 15, 20, 25, 30, 35…these are all going to be composites!
This is actually fairly weird. The primes are scattered in a pattern-free way in the numberline and yet we can build a sequence that included NO primes even though the sequence elements are rather close together!
Popper 10, Question 3
If q = 15 and d = 22, the sequence is prime-free.
A true
B false
Now, still looking at primes, let’s look at an ancient prime-finding device:
The Sieve of Eratosthenes
Eratosthenes was a Greek mathematician who lived quite a bit of time before Christ was born. He invented a foolproof way for students to find primes if they knew their multiplication tables.
Make a list of all natural numbers
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18