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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 25  92 ?
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 = a b c d 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
35 5  7 5


49 7  7 7
110 11

130 13
And you spend time telling them how to cancel common factors…
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Then one of them may show up with this example:
\
26 2 6 2


65 6 5 5
And you’re forced to admit that cancelling the 6’s “works” because
26 2(13) 2


65 5(13) 5
And there are a few others that work like this, too. But it’s a fluke, not a consistent rule.
25 2
 . You can point out that this type of cancelling doesn’t ALWAYS
53 3
Let’s explore the situations in which it DOES work.
If you try it with
work.
First we’ll set up our equation:
xy x y x


yz y z z
Now it’s NOT x times y or y times z in the numerator and denominator!
It’s
10 x  y x
 and x, y, and z are single digits! And the y “cancels wrongly”.
10 y  z z
Popper 09, Question 1
Here’s another one that works with the wrong cancelling:
19 1

95 5
A
True
B
False
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It’s
10 x  y x
 and x, y, and z are single digits! And the y “cancels wrongly”.
10 y  z z
Here’s the plan of attack:
We’ll look at 3 cases:
Case 1
x=y=z
Case 2
x yz
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:
aa a
 1
aa a
Well, yeah:
aa a(11) a

 1
aa a(11) a
However Case 2 offers some interesting work!
Case 2
x yz
Let’s look at our fraction again and do some algebra:
any number divided by itself!
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xy 10 x  y x


yz 10 y  z z
10 x  y x

10 y  z z
z (10 x  y )  x(10 y  z )
10 xz  zy  10 xy  zx
subtract yz and xz from both sides
10 xz  xz  10 xy  yz
9 xz  y (10 x  z )
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
10 x  z 9 0
which means that 10x – z is a multiple of 9
Now:
Add z to both sides:
Note that 10 9 1
10x 9 z
so we’ll replace 10 with 1 in the equivalence:
x 9 z
And we have a contradiction: Case 2 is x  y  z 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
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B.
Suppose 9 divides y
Then 9 = y because these are all single digits!
9 xz  y (10 x  z )
xz  (10 x  z )
Now since 9 = y, we know that neither x nor z is equal to 9.
2B cases
Suppose x = 1, substitute into the equation: xz  (10 x  z )
z  10  z
2 z  10
z 5
x = 1, y = 9, z = 5
xy 19 1


yz 95 5
works!
Suppose x = 2, substitute into the equation: xz  (10 x  z )
2 z  10  2  z
3z  20
Nothing
Suppose x = 3
4z = 30
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Suppose x = 4
4 z  40  z
5 z  40
z 8
Perfect! Another 1:
49 4 1
 
98 3 2
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
16 1

64 4
If x = 2 we get
26 2

65 5
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:
143185
1435

1701856 17056
"cancel the 18" !
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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:
59  14  5
6372  18  10  1
When you find the digital sum of a perfect number greater than 6, you always get 1.
496  19  10  1
8128  19  10  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, Rn , is n the index of the repunit.
A
true
B
false
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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 9 0
Page 22:
A perfect number equals the sum of its proper divisors plus 1.
Page 22:
An even number N is perfect whenever N  2P1 (2 p  1) for P a prime number.
Fact:
All primes greater than 2 are ODD numbers.
Fact:
2P1 is always an even number and we can show the following
2even 3 1
2even  3k  1 k , any integer
while 2odd prime 3 1
Now for N  2P 1 (2P  1)
Now if P is an odd prime:
2 P 1  3k  1 *
So now let’s multiply both sides by 2:
21  2 P 1  2(3k  1)
2 p  6k  2
2 P  1  6k  1**
subtract 1 from both sides
recall [2]=[−1]
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Let’s look back a N, our perfect number:
N  2P1 (2 p  1) and substitute in some algebra formulas for the factors: * and **
N  (3k  1)(6k  1)
N  18k 2  9k  1
N  9(2k 2  1)  1
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
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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 = a 2b 2 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 100ab  10a 2  10b 2  ab .
Now this is not a true statement for larger digit numbers:
169(961)=162409 = 4032
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
10  2  5
zero-free factors
102  22  52
zero-free factors
103  107
zero-free factors
58  390625
109 , and 1018 and 1033
past 105000 ?
zero-free factors
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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 4  52  100 . In fact, every digit place has a cycle that is 4  5n1 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
1
the cycle is computed with  2n1 with n being the digits place from the right.
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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
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Perfect square sums in fractions that are equal to 2:
page 90
a 2  b2
2
c2  d 2
How many solutions are there? LOTS!
Here’s one:
22  42 4  16 20


2
12  32
1  9 10
Here’s one not in the book:
52  7 2 74

2
62  12 37
In the homework you have to find some of these and come up with some tips for finding them.
Adding patterns
Page 91
n(n  1)
1 2  3
456  78
9  10  11  12  13  14  15
...
Let’s take row 2 and see why this works
4 533  7 8
6 = 2(3)
And look at row 3
9 10 11  4  4  4  13 14 15
12 = 3(4)
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Popper 9, Question 5
1(8)  1  9
12(8)  2  98
123(8)  3  987
The next row is:
A
There is no next row – the pattern stops there
B
1234(8)  4  9876
Second video starts here.
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pages 93 – 109
Chapter 8
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 106 ? We don’t like the idea of counting for this,
we want to estimate.
x
where x is the boundary number and “ln x” is a function
ln x
you can find on any scientific calculator – “ln(x)” is the “natural logarithm of x”.
One reasonable estimator is:
Suppose we want the number of primes less than 9
9
 4, 0960765...
ln(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:
15
x
du
log u
0
Li ( x)  
This is an integral and it finds the area under a curve!
Popper 10, Question 1
x
and the fact that ln(12)  2.5, we can say that the number of
ln x
primes less than 12 is approximately:
Given the estimator formula:
A
B
C
4
5
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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
( n  1)! 2
( n  1)! 3
...
( n  1)! ( n  1)
These are ALL composite numbers.
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Example:
Suppose I use n = 5 … n+1 = 6
6! 2  6  5  4  3  2 1  2  2(6  5  4  3 1  1)
This is a composite number
6! 3
6! 4
6! 5
6! 6
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
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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
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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
Ignore 1 – it’s just a place holder. Circle 2 and then cross out all multiples of 3 that follow.
Circle 3 and cross out all multiples of 3 that follow. Skip up to 5, circle it and cross out all
multiples of 5 that follow. Skip to 7, and so forth. The only numbers that end up NOT crossed
off are the primes.
In the book are several elaborations on the classical Sieve. Table 5 on page 97, for example, is
an improvement in that there’s less to write down – all the even numbers are left off in writing
the table … that eliminates 50% of the numbers to write down. You just have to remember that
2 is prime and add it to the list inbetween 1 and 3 when you’re done.
Note that the Table 5 columns are all the ODD mod 10 equivalence classes – there will be no
primes in the even equivalence classes!
On page 98, Table 6 is another sieve – this is a sieve of arithmetic sequences. The clever part
x 1
about this table is if x is an odd natural number greater than 2, then x is prime whenever
is
2
NOT in the table.
The pattern to the table is
The first row has a = 4 and d = 3. Ditto the first column
The second row has a = 7 and d = 5 (and the second column has this, too)
Each subsequent row and column adds 3 to a and 2 to d.
3 1 5 1
,
,... these numbers are NOT in the table. This is because we’ve built
2
2
arithmetic sequences that do NOT contain primes per the instructions earlier in the presentation!
Notice that
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Table 7, page 99, is a list of odd composite numbers (some appear more than once!), but no
prime numbers.
It uses n 
x 1
 2n  1  x, a composite a version of the earlier formula:
2
Note that row 1 contains all multiples of 3 except 3 – and row 2 contains all multiples of 5 except
5…so the gcd of the row is the prime and what’s listed is everything that you’d cross out in the
classic Sieve!
In 1955 at Los Alamos, during a coffee break, the research group led by Stanislaw Ulam visited
a bit about fun math ideas. They started with the Sieve and changed the cross out rules. They
decided that they’d cross out numbers based on POSITION rather than factors.
They made a list of the natural numbers. Circled 1 and crossed out every other number. Then
they went to 3 and circled it and crossed out every third number of the remaining numbers. They
then circled 7 and canceled every 7th number of the remaining numbers. They called the
remaining numbers the Lucky Numbers.
They got a set {1, 3, 7, 9, 13, 15, 21, 25, 31, …} that is shockingly like the Primes!
All of the Luckies are odd (all except one of the primes are odd).
The spaces between the Luckies get further and further apart as you get bigger.
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The estimator for how many Luckies there are less than a given x is
which uses a function
log x
related to the natural logarithm (logarithm base 10).
For the primes there’s a famous conjecture called Goldbach’s Conjecture:
Every even number larger than 2 is expressible as the sum of 2 primes (not necessarily
distinct e.g. 4 = 2 + 2)
There’s a matching conjecture for the Luckies: Every even number is the sum of 2
Luckies (not necessarily distinct e.g. 2 = 1 + 1)
Primes are all primes BUT Luckies are a mixture of primes and composites.
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Popper 10, Question 4
The Sieve of Eratosthenes was invented on a coffee break in 1955.
A
True
B
False
Up until the Luckies, primes were considered to be unique – no other set was quite like
them…but now, the Luckies demonstrate that there may indeed be similar sets!
In fact, even on small details the sets are similar. There are the famous “twin primes” – prime
numbers separated by exactly one composite number (3, 5; 17, 19;…)
And there are “twin Luckies” (7, 9; 13, 15;…)
Both sets of “twins” are infinite in size.
There’s a really cool interactive display on Wikipedia that will let you watch the crossing out
pattern on the Luckies. Be sure to check it out if you have time!
Popper 10, Question 5
Lucky Numbers can be prime or composite or neither.
A
True
B
False
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