Download Pretty Primes Class Notes

Pretty Primes
Introduction
This is a cycle of 2-3 lessons looking at primes numbers and the prime factorisation of
numbers.
Resources
 Calculators for students will be useful for students dealing with larger numbers.
 A copy of the Question Sheet, one for each student.
Sections
1. Locker Games
2. Counting divisors in pairs
3. Unique Prime Factorization
4. Counting divisors, Prime Factorisation: Applications
5. Divisor rule: Applications
6. The Power of Products: Applications
7. Panicky Primes or the Return of the Unique Prime Factorisation
8. Checking for primes: a shortcut.
9. An Important Property of Primes: Applications
1
Pretty Primes - Class Notes
1. Locker Games
One day Anna brings a big roll of stickers to school and she
and her classmates decide to play a game with their lockers.
The first student puts a sticker on every locker. The second
student puts a sticker on the second locker and then on
every second locker after that. The third student then
decorates the third locker and every third locker after that,
and so on, until all 20 students are done decorating. They
tally the number of stickers on each locker.
a) Write each locker number and the numbers of stickers on it in a table. Do you notice
any interesting patterns?
b) What does the number of stickers tell us about the corresponding locker number?
c) Do you notice anything special about the lockers with 2 stickers? Explain.
d) There are very few lockers with odd numbers of stickers. Do you notice anything
special about them? Can you guess a general rule from here? Explain why the rule
should be true.
e) Look at the lockers 1, 2, 4, 8, 16. Anything special about their sticker numbers? Can
you guess a general rule from here? Can you prove it?
f) Do you notice any other patterns?
2. Counting divisors in pairs.
Next Anna and her classmates thought how cool it would be it they had 3,529,147
lockers in their school ... think of how many stickers they would get to use on the last
one! ... In truth, they had no idea how many.
Goal:
Find a systematic way to calculate the number of divisors of any number.
Let’s start small. Take 48 as an example.
Method I for counting divisors:
a) Find all pairs of numbers whose product is 48.
For this we sift through numbers in increasing order, starting from 1. Once we find a
divisor of 48, we write 48 as a product between that divisor and a second number. The
second factor will be a divisor of 48 as well. Through how many numbers do we need to
sift before we’re sure we’ve exhausted all possibilities?
b) Replace 48 by an unknown number 𝑛. Through how many numbers do we need to
sift before we know we’ve exhausted all ways of writing 𝑛 as a product? Write your
answer in terms of 𝑛.
3. Unique Prime Factorization
The first method can be very slow. To quicken it, let’s put our engineer caps on: we’ll
first break our number into smaller parts and then construct our divisors out of the
available components:
2
Method II for counting divisors:
a) Write 48 as a product of prime numbers: This is called prime factorising 48. In how
many ways can you do it, if the order of factors inside the product doesn’t matter?
b) Let 𝑑 be any random divisor of 48.
i) What prime factors can 𝑑 have?
ii) How often can each prime come up as factor when you prime factorise 𝑑?
c) Use this to calculate how many divisors 48 has.
d) Repeat the same for each of these numbers:
𝑖) 28,
𝑖𝑖) 100
𝑖𝑖𝑖) 60.
Check your answer by listing all divisors.
4. Counting divisors, Prime Factorisation: Applications
a) Remember the rule we found in the Locker Games:
The square numbers are the only numbers with an odd number of divisors.
i) What can you say about the indices of all primes in the prime factorisation of a
square number?
ii) Can you use your observations on indices to write a new proof of the rule above?
b) Can you find a rule about the number of divisors of a cube number?
Hint: It may involve the remainder when dividing this number by 3.
c) Find the smallest positive natural number whose product with 2016 is a square
number.
5. Divisor rule: Applications
We can also use the rule below to find out if a number b is divisible by a given number a.
Divisor rule: Given two numbers 𝑎 and 𝑏,
𝑎 divides 𝑏 if and only if the primes of 𝑎 are among the primes of 𝑏,
and their indices are at most equal to the indices in the prime factorisation of 𝑏.
a) i) Prove that the product of any three consecutive numbers is divisible by 6.
ii) For any natural number n, prove that 𝑛(𝑛 + 1)(2𝑛 + 1) is divisible by 6.
b) Find all possible pair of digits x and y such that 36 divides the 4-digit number with
digits x95y.
6. The Power of Products: Applications
Physicists and Chemists always wanted to know what stuff is
made of: they found molecules, atoms, electrons, quarks,
kept together by gravity, electromagnetic, strong or weak
forces...
We wanted to know what numbers are made of, well, we found:
Every natural number is constructed out of its own set of primes with indices, “glued
together” by multiplication.
3
This rule is so powerful, that most of the problems where you have to find some natural
numbers satisfying some special conditions are solved based on products and their
close relatives: divisibility and prime factorisation.
That’s because products are the key to the very structure of natural numbers.
Example:
Find all possible values of a square number with the property that, if you add 15 to it,
the result is still a square number.
7. The Return of the Unique Prime Factorisation, or Panicky Primes
We return to the rule we discovered:
The Unique Prime Factorisation: If we don’t mind the order of the factors,
Any positive natural number can be written in a unique way as a product of primes with
indices.
Class survey/discussion:
a) Think about the time when we drew different factor trees for 48 and we always found
the same “hanging fruits” with same indices. Did this surprise you?
If yes, why? If not, why not?
b) Think about the Unique Prime Factorisation rule. Are you convinced that it’s true
for all natural numbers? If yes, why? If not, why not?
c) As the name suggests, the Unique Prime Factorisation is due to the very special
nature of prime numbers. Let’s reformulate this rule in terms of primes:
4
Suppose that we write a number as a product 𝑎𝑏, and start a factor tree like this:
What does the Unique Prime
Factorisation tell us about the prime
factors of 𝑎𝑏?
(Hint: think in terms of “hanging fruit”)
8. Checking for primes: a shortcut.
We saw before that it’s important to be able to check if a number is prime as quickly as
possible.
Example:
a) Is 2011 a prime?
To find factors of 2011 in pairs, we saw before that it’s enough to check all the numbers
smaller than 45 > √2011. But is it important to check all these numbers, or only some
of them?
b) What are the prime factors of 3,529,147 ? (not really!)
A Tip for Maths Competitors:
Before you go into a maths competition, it may be a good idea to prime factorise the
current year number. If you’re in a prime year, chances are you’re going to use this in a
problem .
9. An Important Property of Primes: Applications
Let 𝑝 be a prime. If 𝑝 | 𝑎𝑏, then 𝑝 | 𝑎 or 𝑝| 𝑏.
a) Find all natural numbers 𝑛 such that 2𝑛 + 1 is a multiple of 7.
b) Your friend has chosen a five digit mystery number. She adds twice the number
formed by the last two digits of the mystery number with three times the number
formed by the first three digits (read from left to right). The result is divisible by 197.
Prove that the mystery number is also divisible by 197.
For example, suppose,𝑛 =
462 95 .
2 × 95 + 3 × 462 = 1576 = 197 × 8. You can also check that 46295 = 197 × 235.
c) In the question b) above, is it important that 𝑛 has 5 digits? Can you formulate a
similar problem that works for numbers 𝑛 with any number of digits?
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Pretty Primes – Practice Questions
1. A class of 50 has been playing the locker game.
a) How many lockers have exactly 2 stickers each?
b) How many lockers have an odd number of stickers each?
c) Which lockers have exactly 3 stickers each?
d) Find a sequence of 6 lockers with consecutive numbers of stickers.
e) Find a sequence of 6 lockers with consecutive numbers of stickers under the
condition that, with the exception of the first student, all other students placed a
sticker on at most 1 of the 6 lockers.
2. How many divisors do the following numbers have?
𝑎)𝑖) 118
𝑖𝑖)24 × 36
𝑖𝑖𝑖) 55 × 78
𝑖𝑣) 1113 × 1311 𝑣) 3 × 1717
𝑏) 𝑖) 98
𝑖𝑖) 108
𝑖𝑖𝑖)256
𝑖𝑣)1000
𝑣) 1,000,000
𝑐)𝑖) 62 × 63 × 611
𝑖𝑖)610 × 1015 × 156
𝑖𝑖𝑖)10 × 9 × 8 × … × 2 × 1
3. One day, every student in the galaxy gets bored and decides to play the locker game.
The 1010 students collect 1010 lockers for the game.
a) How many stickers will there be on the last locker?
b) Among the first 100 lockers, which have exactly 7 stickers each?
c) Among the first 100 lockers, which have exactly 12 stickers each?
5. a) Prove that the product of any 5 consecutive numbers is divisible by 120.
b) Find all possible digits 𝑎 and 𝑏 for which to the 4-digit number with digits 2𝑎9𝑏 is
divisible by 12.
c) Find all possible digits 𝑎 and 𝑏 for which 2𝑎 × 9𝑏 is equal to the 4-digit number
with digits 2𝑎9𝑏.
d) A 5 digit mystery number has the property that the sum of its first, third and fifth
digit is 19, while the sum of the second and third digit is 8. Prove that the number has at
least 24 divisors.
Hint: Check divisibility by 9 and by 11 first. Can you prove that the number has 12
divisors? Where could the other 12 come from?
6. a) Find all possible values of a square number with the property that, if you add 12 to
it, the result is still a square number.
b) “There exists no square number with the property that, if you add 𝑛 to it, the result
is still a square number.”
i) Give 2 examples of natural numbers 𝑛 for which this statement is true.
ii) Can you describe all natural numbers 𝑛 for which this statement is true?
c) Find all natural numbers 𝑥 for which 1 + 99 × 2𝑥 is the square of a natural number.
d) Find the natural numbers 𝑥, 𝑦, 𝑧 , all larger than 1, knowing that
3𝑥 + 3𝑦 + 3𝑧 + 3𝑡 = 6840.
7. a) Prove the tongue-twister:
The divisor of a divisor of a number is also a divisor of the number.
b) Here is an example of three natural numbers such that the first divides the product
of the second and third, but it divides neither the second, nor the third number: 6
divides (4 × 9) but 6 doesn’t divide 4, and 6 doesn’t divide 9.
Find other examples.
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c) Recall the Property of Primes: Let 𝑝 be a prime. If 𝑝 | 𝑎𝑏, then 𝑝 | 𝑎 or 𝑝| 𝑏.
Let 𝑝 and 𝑞 be two mystery prime numbers with the property that 𝑞10 is a multiple of 𝑝.
Prove that 𝑞 and 𝑝 are equal.
8. a) Check if any of these numbers are prime:
𝑖) 2011 𝑖𝑖)2013
𝑖𝑖𝑖)2015
𝑖𝑣)2017 𝑣) 2019.
9. a) Find all natural numbers 𝑛 such that 3𝑛 + 1 is a multiple of 5.
10. a) A 7 digit number is called unlucky if twice the number formed by its last two
digits plus five times the number formed by its first five digits is a multiple of 13.
Prove that an unlucky number is also divisible by 13.
Pretty Primes - Class Notes
1. Locker Games
7
This game can be played by the whole class
even if you don’t have lockers nearby.
Partition the blackboard into 20 vertical
regions and label them by consecutive
numbers 1—20. These are your lockers.
One day Anna brings a big roll of stickers to school and she and her classmates decide to
play a game with their lockers. The first student puts a sticker on every locker. The
second student puts a sticker on the second locker and then on every second locker
after that. The third student then decorates the third locker and every third locker after
that, and so on, until all 20 students are done decorating. They tally the number of
stickers on each locker.
a) Do the tally yourselves and write the Locker and Sticker numbers in a table. Do you
notice any interesting patterns?
b) What does the number of stickers tell us about the corresponding locker number?
c) Do you notice anything special about the lockers with 2 stickers? Explain.
d) There are very few odd numbers of stickers. Do you notice anything special about
their lockers? Can you guess a general rule from here? Can you explain why the rule
should be true?
e) Look at the lockers 1, 2, 4, 8, 16. Anything special about their sticker numbers? Can
you guess a general rule from here? Can you prove it?
f) Do you notice any other patterns? Try various ideas.
Solutions: a)
Locker
Stickers
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1 2 2 3 2 4 2 4 3 4 2 6 2 4 4 5 2 6 2 6
b) Hint: Think about how each locker got its stickers.
Answer: There was one sticker added for each divisor (factor) of the locker number.
The stickers count the number of divisors (factors) of the locker number.
c) The lockers with 2 stickers represent the numbers with exactly two divisors.
These are known as the prime numbers.
Prime numbers are numbers divisible only by 1 and themselves.
d) 1, 4, 9, 16 are all the square numbers between 1 and 20. We make the following
guess:
The square numbers are the only numbers with an odd number of divisors.
Hint for proof: If you can’t prove the rule right away, first try it out on 36. What’s a fast
way to write all its divisors?
Proof: Write 36 as product of two numbers in all possible ways.
8
36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6.
The factors are all the divisors of 36. They all come paired with a different factor
except for 6 which is paired with itself exactly because 36 is the square of 6. The same
argument works for all square numbers.
By contrast, if a number is not a square number, each of its factors will come paired
with a different factor from itself.
e)
Number
No. of divisors
1 = 20
1
2 = 21
2
4 = 22
3
8 = 23
4
16 = 24
5
Writing 1 = 20 , 2 = 21 , 4 = 22 , 8 = 23 , 16 = 24 we see that the sticker numbers
are 1 unit larger than the powers (indices) of 2 in each locker number.
Why is that? Let’s look at the divisors of 8:
1 = 20 , 2 = 21 , 4 = 22 , 8 = 23 .
We can count them by their indices but we have to include 0. Hence, the number of
divisors of 2𝑛 is 𝑛 + 1.
f) E.g. The same pattern holds for 3𝑛 : it has 𝑛 + 1 divisors.
2. Counting divisors in pairs.
Next Anna and her classmates thought how cool it would be it they had 3,529,147
lockers in their school ... think of how many stickers they would get to use on the last
one! ... In truth, they had no idea how many.
Goal:
Find a systematic way to calculate the number of divisors of any number.
Let’s start small. Take 48 as an example.
Exercise: Method I for counting divisors:
a) Find all pairs of numbers whose product is 48.
For this we sift through numbers in increasing order, starting from 1. Once we find a
divisor of 48, we write 48 as a product between that divisor and a second number. The
second factor will be a divisor of 48 as well. Through how many numbers do we need to
sift before we’re sure we’ve exhausted all possibilities?
b) Replace 48 by an unknown number 𝑛. Through how many numbers do we need to
sift before we know we’ve exhausted all ways of writing 𝑛 as a product? Write your
answer in terms of 𝑛.
Solutions:
a) 48 = 1 × 48 = 2 × 24 = 3 × 16 = 4 × 12 = 6 × 8.
As the first factor in the products increases, the second factor decreases. We have to
stop at 6 × 8 because the next first factor must be at least 7, while the second factor
must be at most 7, but 7 × 7 = 49. After that, the second factor would get to be less than
7, but we have already found all divisors less than 7.
b) By the same logic, we have to check numbers 1 through the closest natural number to
√𝑛.
9
Conclusions:
To find all the divisors of 3,529,147 by this method, we’d have to sift through
√3,529,147 numbers. That’s not too far from √4,000,000 = 2,000 checks. A bit much!
3. Unique Prime Factorization
The first method can be very slow. To quicken it, let’s put our engineer caps on: we’ll
first break our number into smaller parts and then construct our divisors out of the
available components:
Exercise: Method II for counting divisors:
a) Write 48 as a product of (possibly more than two) prime numbers: This is called
prime factorising 48. In how many ways can you do it?
b) Let 𝑑 be any random divisor of 48.
i) What prime factors can 𝑑 have?
ii) How often can each prime come up as factor when you prime factorise 𝑑?
c) Use this to calculate how many divisors 48 has.
Solutions:
a) Let’s write our number as a product.
48 = 8 × 6.
But 6 and 8 have their own divisors: 8 = 2 × 4 and 6 = 3 × 2. And 4 has its own
divisors: 4 = 2 × 2. We put the results up in a tree.
This is referred to as a
factor tree of 48.
The prime factors of 48 are the
“fruits” at the end of the branches.
2 comes up 4 times and 3 once.
All in all 48 = 6 × 8 = 3 × 2 × 4 × 2 = 3 × 2 × 2 × 2 × 2 = 3 × 24 = 24 × 3.
(Note: no brackets are necessary because all operations are multiplications).
We can’t factorise anymore, because 2 and 3 are prime. When we have expressed a
number as a product of primes, the number is prime factorised.
But we can write 48 as a product is a number of ways. So different people can find
different factor trees of 48. Let’s check our various examples and compare the results:
10
There are other trees, but happily, they all have the same “hanging fruits”, so they all
result in the same prime factorisation
48 = 3 × 24 = 24 × 3.
If we don’t mind the order of the factors we can say that 48 has a unique prime
factorization.
b) Let 𝑑 be a divisor of 48.
i) The prime factors of d have to be among the prime factors of 48, that is 2 and 3.
ii) Each of them will appear in the prime factorisation of 𝑑 at most as often as it appears
in the prime factorisation of 48, that is:
 2 will come up at most 4 times, that is 0, 1, 2, 3 or 4 times
 3 will come up at most once that is 0 or 1 time.
We can see this from the factor trees as follows: each divisor of 48 is a branch in a
factor tree. Its prime factors, the “fruits” of the branch, are also fruits of the tree.
We can prove this more rigorously using this tongue-twister:
The divisor of a divisor of a number is also a divisor of the number.
as well as the rule of division with indices. We’ll leave this proof as an exercise.
c) We write all divisors of 48 as product of primes and arrange them in a table
according to the indices of 2 and 3:
1 = 20 × 30
2 = 21 × 30
4 = 22 × 30
8 = 23 × 30
16 = 24 × 30
3 = 20 × 31
6 = 21 × 31
12 = 22 × 31
24 = 23 × 31
48 = 24 × 31
11
There are 5 choices of indices for 2. Each of these can be paired with any of the 2 choices
of indices of 3, so the result is a total of 5 × 2 = 10 choices.
Conclusions:
a) Unique Prime Factorisation: If we don’t mind the order of the factors,
Any positive natural number can be written in a unique way as a product of primes with
indices.
b) Divisor rule: Given two numbers 𝑎 and 𝑏,
𝑎 divides 𝑏 if and only if the prime divisors of 𝑎 , with their given indices, divide 𝑏.
Example: 24 divides b if 8 divides b and 3 divides b.
c) Counting divisors: Consider a number 𝑎, prime-factorised.
To find how many divisors 𝑎 has, add 1 to each index in the prime-factorisation, then
multiply the results.
Practice exercises:
For each of these numbers:
𝑖) 28,
𝑖𝑖) 100
𝑖𝑖𝑖) 60.
Prime-factorise them; find how many divisors they have and check your answer by
listing all divisors.
4. Counting divisors, Prime Factorisation: Applications
Exercise:
a) Remember the rule we found in the Locker Games:
The square numbers are the only numbers with an odd number of divisors.
Let’s look at the prime factorisation of a square number:
i) What can you say about the indices of all primes in the prime factorisation of a
square number?
ii) Can you use your observations on indices to write a new proof of the rule above?
b) Can you find a rule about the number of divisors of a cube number?
Hint: It may involve the remainder when dividing this number by 3.
c) Find the smallest positive natural number whose product with 2016 is a square
number.
Solutions:
a) i) All the indices have to be even, that is because a square number is written as 𝑥 2 .
The primes divisors of 𝑥 2 are the same as the primes of 𝑥, but their indices are doubled.
ii) Since every prime in the prime factorisation of 𝑥 2 has an even index, when we add 1
we get an odd number. Multiplying all these odd numbers gives an odd number. This is
the number of divisors of 𝑥 2 .
12
On the other hand, suppose that our number is not a square number. This means that
one of the primes in its prime factorisation comes with an odd index. Add 1 and you get
an even number. Multiply it by anything and the result is always even.
b) The number of divisors of a cube number always yields remainder 1 when divided by
3. That is because all its primes come with indices which are multiples of 3.
c) 2016 = 25 × 32 × 71 . When we multiply 2016 by another number, we can either
increase the indices of the primes 2, 3, 7, or bring in new primes. As we’re looking for
the smallest product, it makes sense to only modify indices so that they all become even:
5 becomes 6; 2 stays 2 and 1 becomes 2. For this, we multiply 2016 by 2 × 7 = 14 and
we get
6
2
2016 × 14 = 25 × 32 × 71 × 2 × 7 = 2 × 32 × 72 = (23 × 3 × 7) .
5. Divisor rule: Applications
We can also use the rule below to find out if a number b is divisible by a given number a.
Divisor rule: Given two numbers 𝑎 and 𝑏,
𝑎 divides 𝑏 if and only if the primes of 𝑎 are among the primes of 𝑏,
and their indices are at most equal to the indices in the prime factorisation of 𝑏.
In this case we know the prime factorisation of the first number a. It remains to check if
the second number b has those same prime factors, with at least the indices they have in
a.
Example:
a) i) Prove that the product of any three consecutive numbers is divisible by 6.
ii) For any natural number n, prove that 𝑛(𝑛 + 1)(2𝑛 + 1) is divisible by 6.
Solution:
The prime factorisation of 6 is 6 = 2 × 3. By the Divisor rule,
A number is divisible by 6 if it is divisible by 2 and by 3.
i) Among any three consecutive numbers there is at least one even number, and at least
one number which is a multiple of 3. So their product will be divisible by both 2 and 3.
The End.
ii) By the same logic as above, one of the numbers 𝑛 , (𝑛 + 1) is even. It remains to
prove that among 𝑛, (𝑛 + 1), (2𝑛 + 1) there is a number which is a multiple of 3. Let’s
see what happens if neither 𝑛 nor 𝑛 + 1 is a multiple of 3. Then 𝑛 + 2 must be. So 𝑛 +
2 = 3𝑘 for some natural number 𝑘, or equivalently, 𝑛 = 3𝑘 − 2.
Let’s see what this says about (2𝑛 + 1):
2𝑛 + 1 = 2(3𝑘 − 2) + 1 = 6𝑘 − 4 + 1 = 6𝑘 − 3 = 3(2𝑘 − 1).
Example:
13
b) Find all possible pair of digits x and y such that 36 divides the 4-digit number with
digits x95y.
Solution:
The prime factorisation of 36 is 36 = 22 × 32 . By the Divisor rule,
A number is divisible by 36 iff it is divisible by 22 = 4 and by 32 = 9.
The Divisibility by 4 rule implies that 4| 𝑥95𝑦 iff 4| (50 + 𝑦).
As 50 + 𝑦 = 48 + 2 + 𝑦, we get the cases: i) 𝑦 = 2; and ii) 𝑦 = 6.
The Divisibility by 9 rule implies that 9| 𝑥95𝑦 iff 9| (𝑥 + 9 + 5 + 𝑦).
Case i) 𝑦 = 2 and 𝑥 + 9 + 5 + 2 = 𝑥 + 16 so 𝑥 = 2.
ii) ) 𝑦 = 6 and 𝑥 + 9 + 5 + 6 = 𝑥 + 20 so 𝑥 = 7.
6. The Power of Products: Applications
Physicists and Chemists always wanted to know what stuff is
made of: they found molecules, atoms, electrons, bosons and
fermions, kept together by gravity, electromagnetic, strong or
weak forces...
If we wanted to know what numbers are made of, well, we found:
Every natural number is constructed out of its own set of primes with indices, “glued
together” by multiplication.
This rule is so powerful, that most of the problems where you have to find some natural
numbers satisfying some special conditions are solved based on products and their
close relatives: divisibility and prime factorisation.
That’s because products are the key to the very structure of natural numbers.
Example:
Find all possible values of a square number with the property that, if you add 15 to it,
the result is still a square number.
Solution:
Let’s give the unknown number a name. Since it’s a square number, we can call it 𝑥 2 .
When we add 15 to it we obtain another square number, let’s call it 𝑦 2 . So
𝑦 2 = 𝑥 2 + 15.
It’s impossible to find all solutions of this equation by trial and error: 𝑥 could a very big
number, and then so could 𝑦.
What does this equation have to do with products?
We could factorise 15 = 3 × 5. Then let’s get 15 all on its own on the right hand side of
the equation:
𝑦 2 = 𝑥 2 + 15
becomes
𝑦 2 − 𝑥 2 = 15.
Do you notice/remember anything special about 𝑦 2 − 𝑥 2 ?
In the Distributivity class we discussed a nice trick based on the formula
14
𝑦 2 − 𝑥 2 = (𝑦 − 𝑥)(𝑦 + 𝑥).
So our equation now becomes (𝑦 − 𝑥)(𝑦 + 𝑥) = 15 = 3 × 5.
At this point we could identify
the smaller factors on each side as 𝑦 − 𝑥 = 3,
and the larger factors as
𝑦+𝑥 =5.
Add up the terms to get 2𝑦 = 8 so 𝑦 = 4, and then 𝑥 = 5 − 4 = 1.
So our solution is, not surprisingly, 𝑥 2 = 1.
To check our answer, we see that 1+15=16 is indeed a square number.
But is this the only solution?
We should remember that 15 = 1 × 15 as well so another option is
𝑦 − 𝑥 = 1,
𝑦 + 𝑥 = 15 .
Add up the terms to get 2𝑦 = 16 so 𝑦 = 8, and then 𝑥 = 15 − 8 = 7.
Check: 49+15=64, a square number.
Problem solvers beware:
Some quick problem solvers may have found the solutions 1 and 49 with a few trials.
Will they get the big prize for their solution?
Answer: No, not from mathematicians for sure. When we solve a problem, we want to
be absolutely sure that we have found all solutions. Think about clearing two land mines
from a minefield... wouldn’t you like to know that there are absolutely no other land
mines hidden around?
This last step is worth all the money in mathematics, because in this case you’re
eliminating infinitely many hidden possibilities – pretty hard job if you ask me!
7. The Return of the Unique Prime Factorisation, or Panicky Primes
We return to the rule we discovered:
The Unique Prime Factorisation: If we don’t mind the order of the factors,
Any positive natural number can be written in a unique way as a product of primes with
indices.
Class survey/discussion:
a) Think about the time when we drew different factor trees for 48 and we always found
the same “hanging fruits” with same indices. Did this surprise you?
If yes, why? If not, why not?
15
b) Think about the Unique Prime Factorisation rule. Are you convinced that it’s true
for all natural numbers? If yes, why? If not, why not?
c) As the name suggests, the Unique Prime Factorisation is due to the very special
nature of prime numbers. Let’s reformulate this rule in terms of primes:
Suppose that we write a number as a product 𝑎𝑏, and start a factor tree like this:
What does the Unique Prime Factorisation tell us about the
prime factors of 𝑎𝑏? (Hint: think in terms of “hanging fruit”)
Moderator’s Comments:
a) Some people may not have been surprised by this discovery. This may be because
they already knew all the factors of 48 and could quickly check that they have no prime
factors other than 2 and 3. Or they may have noticed that 48 could not possibly have any
prime factors other than 2 and 3: indeed, the only prime they needed to check would be
5, since 7 already satisfies
7 × 7 = 49 > 48.
These checks can go on so quickly in our minds that we may find the results absolutely
natural and obvious...
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b) Even though a rule holds for 48 and other numbers small enough to check, this does
not guarantee that the rule will be true for all numbers.
Suppose that we choose a really, really large number, so that our
lives won’t suffice to find all its factors.
Suppose that we manage to find a factor tree for that number.
(This is an easier task, since it’s enough to find two factors, and then two factors
of the factors, etc. The numbers decrease with each step.)
What guarantee do we have that there are no primes outside the
tree, but which divide the number – like some fruits not hanging from
the branches?
(They might be part of another factor tree of the same number!)
c)
We can reformulate the Unique Prime Factorisation rule like this:
If 𝑝 is a fruit of 𝑎𝑏, then it hangs from one of its branches 𝑎 or 𝑏!
More Mathematically put:
Important Property of Primes!
Let 𝑝 be a prime. If 𝑝 | 𝑎𝑏, then 𝑝 | 𝑎 or 𝑝| 𝑏.
Conclusions:
The following scary questions follow directly from our discussion:
d) Is Unique Prime Factorisation true even for very large numbers, with very large
prime factors?
e) How large can prime numbers get anyway? Are there infinitely many primes?
f) We reformulated the Unique Prime Factorisation rule as a property of primes:
If 𝑝 | 𝑎𝑏, then 𝑝 | 𝑎 or 𝑝| 𝑏.
Is this property true for all primes? Where does it come from? Can we prove it?
You can start panicking now!
Answers:
Either: you can think about these matters for a while, and we’ll discuss them later. Or:
e) There are indeed infinitely many primes, and we’ll prove this in a future lesson not
too far from now.
f) Is also true, and it is due to way we divide natural numbers with remainder, as well as
from the definition of primes. If you’re really really curious, we will look at a proof in a
future lesson, after we discuss the gcd (greatest common divisor, or highest common
factor) of two numbers.
Because f) is true, then d) is true as well.
8. Checking for primes: a shortcut.
We saw before that it’s important to be able to check if a number is prime as quickly as
possible.
Example:
a) Is 2011 a prime?
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To find factors of 2011 in pairs, we saw before that it’s enough to check all the numbers
smaller than 45 > √2011. But is it important to check all these numbers, or only some
of them?
b) What are the prime factors of 3,529,147 ? (not really!)
Answers:
a) No. It’s enough to check if any of the primes less than 45 are factors of 2011. Indeed,
if 2011 has some factors other than 1 and itself, then it automatically has some prime
factors other than 1 and itself!
You need to check if these numbers divide 2011:
2
3
5
7
11
13
17
19
23
29
31
37
41
43
b) List of prime numbers smaller than 2000. If you’re really patient and have nothing
else to do, you can use these and a calculator to check that 3,529,147 is prime.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
21–40
73
79
83
89
97
101 103 107 109 113 127 131 137 139 149 151 157 163 167
173
41–60
179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277
281
61–80
283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401
409
81–100 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523
541
101–120 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653
659
121–140 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797
809
141–160 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937
941
1–20
161–180 947 953 967 971 977 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069
181–200 1087 1091 1093 1097 1103 1109 1117 1123 1129 1151 1153 1163 1171 1181 1187 1193 1201 1213 1217 1223
201–220 1229 1231 1237 1249 1259 1277 1279 1283 1289 1291 1297 1301 1303 1307 1319 1321 1327 1361 1367 1373
221–240 1381 1399 1409 1423 1427 1429 1433 1439 1447 1451 1453 1459 1471 1481 1483 1487 1489 1493 1499 1511
241–260 1523 1531 1543 1549 1553 1559 1567 1571 1579 1583 1597 1601 1607 1609 1613 1619 1621 1627 1637 1657
261–280 1663 1667 1669 1693 1697 1699 1709 1721 1723 1733 1741 1747 1753 1759 1777 1783 1787 1789 1801 1811
281–300 1823 1831 1847 1861 1867 1871 1873 1877 1879 1889 1901 1907 1913 1931 1933 1949 1951 1973 1979 1987
A Tip for Maths Competitors:
Before you go into a maths competition, it may be a good idea to prime factorise the
current year number. If you’re in a prime year, chances are you’re going to use this in a
problem .
9. An Important Property of Primes: Applications
Let 𝑝 be a prime. If 𝑝 | 𝑎𝑏, then 𝑝 | 𝑎 or 𝑝| 𝑏.
Example:
a) Find all natural numbers 𝑛 such that 2𝑛 + 1 is a multiple of 7.
Solution:
A quick check will yield 𝑛 = 3, 10, 17, 24, …
We start to guess a pattern: we start from 3 and go every 7th number after that.
We can write such numbers as 𝑛 = 3 + 7𝑘, where 𝑘 can be any natural number.
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But how can we check/prove that these are all the solutions?
One natural first step would be to subtract 3 from our number 𝑛. This will mean
subtracting 6 from 2𝑛, so
2𝑛 + 1 = 2𝑛 − 6 + 7 = 2(𝑛 − 3) + 7.
Now 7 already divides 7, so 7 divides 2𝑛 + 1 if and only if 7 divides 2(𝑛 − 3).
By the property of primes, if 7 divides 2(𝑛 − 3) then 7 divides 2 or 𝑛 − 3. Clearly 7
doesn’t divide 2, so it must divide 𝑛 − 3. In other words, 𝑛 − 3 = 7𝑘, where 𝑘 can be
any natural number.
Exercise:
b) Your friend has chosen a five digit mystery number. She adds twice the number
formed by the last two digits of the mystery number with three times the number
formed by the first three digits (read from left to right). The result is divisible by 197.
Prove that the mystery number is also divisible by 197.
For example, suppose,𝑛 = 462 95 .
2 × 95 + 3 × 462 = 1576 = 197 × 8. You can also check that 46295 = 197 × 235.
c) In the question b) above, is it important that 𝑛 has 5 digits? Can you formulate a
similar problem that works for numbers 𝑛 with any number of digits?
Solution:
Let’s denote the number formed by the last two digits by 𝑥 and the number formed by
the first three digits by 𝑦. How can we write the mystery number 𝑛 as a formula in 𝑥 and
𝑦?
For example, 46295 = 46200 + 95 = 95 + 100 × 462 .
In general,
𝑛 = 𝑥 + 100𝑦.
Now the operations described in the problem are 2𝑥 + 3𝑦.
We know that 197 divides 2𝑥 + 3𝑦 and we’d like to prove that 197 divides 𝑥 + 100𝑦.
Since in the first number, 𝑥 comes up twice, it’s natural to look at
2𝑛 = 2(𝑥 + 100𝑦) = 2𝑥 + 200𝑦. We compare to 2𝑥 + 3𝑦:
2𝑛 = 2𝑥 + 200𝑦 = 2𝑥 + 3𝑦 + 197𝑦. Now we know that 197| (2𝑥 + 3𝑦) and 197|197𝑦,
so 197| (2𝑥 + 200𝑦) = 2𝑛. Since 197 doesn’t divide 2, it must be that 197 divides 𝑛.
c) My friend has chosen a mystery number. She adds twice the number formed by the
last two digits with three times the number formed by the remaining digits (read from
left to right). The result is divisible by 197.
Prove that the mystery number is also divisible by 197.
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Pretty Primes – Practice Questions
4. A class of 50 has been playing the locker game.
a) How many lockers have exactly 2 stickers each?
b) How many lockers have an odd number of stickers each?
c) Which lockers have exactly 3 stickers each?
d) Find a sequence of 6 lockers with consecutive numbers of stickers.
e) Find a sequence of 6 lockers with consecutive numbers of stickers under the
condition that, with the exception of the first student, all other students placed a
sticker on at most 1 of the 6 lockers.
5. How many divisors do the following numbers have?
𝑎) 𝑖) 210
𝑖𝑖) 118
𝑖𝑖𝑖) 1731
𝑏) 𝑖)24 × 36
𝑖𝑖) 55 × 78
𝑖𝑖𝑖) 1113 × 1311 𝑖𝑣) 3 × 1717
𝑐) 𝑖) 98
𝑖𝑖) 108
𝑖𝑖𝑖)256
𝑖𝑣)1000
𝑣) 1,000,000
2
3
11
10
15
6
𝑑)𝑖)6 × 6 × 6
𝑖𝑖)6 × 10 × 15
𝑖𝑖𝑖)10 × 9 × 8 × … × 2 × 1
6. One day, every student in the galaxy gets bored and decides to play the locker game.
The 1010 students collect 1010 lockers for the game.
a) How many stickers will there be on the last locker?
b) Among the first 100 lockers, which have exactly 7 stickers each?
c) Among the first 100 lockers, which have exactly 12 stickers each?
5. a) Prove that the product of any 5 consecutive numbers is divisible by 120.
b) Find all possible digits 𝑎 and 𝑏 for which to the 4-digit number with digits 2𝑎9𝑏 is
divisible by 12.
c) Find all possible digits 𝑎 and 𝑏 for which 2𝑎 × 9𝑏 is equal to the 4-digit number
with digits 2𝑎9𝑏.
d) A 5 digit mystery number has the property that the sum of its first, third and fifth
digit is 19, while the sum of the second and third digit is 8. Prove that the number has at
least 24 divisors.
Hint: Check divisibility by 9 and by 11 first. Can you prove that the number has 12
divisors? Where could the other 12 come from?
6. a) Find all possible values of a square number with the property that, if you add 12 to
it, the result is still a square number.
b) “There exists no square number with the property that, if you add 𝑛 to it, the result
is still a square number.”
i) Give 2 examples of natural numbers 𝑛 for which this statement is true.
ii) Can you describe all natural numbers 𝑛 for which this statement is true?
c) Find all natural numbers 𝑥 for which 1 + 99 × 2𝑥 is the square of a natural
number.
d) Find the natural numbers 𝑥, 𝑦, 𝑧 and , all larger than 1, knowing that
3𝑥 + 3𝑦 + 3𝑧 + 3𝑡 = 6840.
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7. a) Prove the tongue-twister:
The divisor of a divisor of a number is also a divisor of the number.
b) Here is an example of three natural numbers such that the first divides the product
of the second and third, but it divides neither the second, nor the third number: 6
divides (4 × 9) but 6 doesn’t divide 4, and 6 doesn’t divide 9.
Find other examples.
c) Recall the Property of Primes: Let 𝑝 be a prime. If 𝑝 | 𝑎𝑏, then 𝑝 | 𝑎 or 𝑝| 𝑏.
Let 𝑝 and 𝑞 be two mystery prime numbers with the property that 𝑞10 is a multiple of 𝑝.
Prove that 𝑞 and 𝑝 are equal.
8. a) Check if any of these numbers are prime:
𝑖) 2011 𝑖𝑖)2013
𝑖𝑖𝑖)2015
𝑖𝑣)2017 𝑣) 2019.
9. a) Find all natural numbers 𝑛 such that 3𝑛 + 1 is a multiple of 5.
10. a) A 7 digit number is called unlucky if twice the number formed by its last two
digits plus five times the number formed by its first five digits is a multiple of 13.
Prove that an unlucky number is also divisible by 13.
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Pretty Primes - Activity Sheet Solutions
1. a) 15 prime numbers are smaller than 50.
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
b) There are 6 square numbers smaller than 50, namely 4, 9, 16, 25, 36, 49.
c) 4, 9, 25, 49: the squares of primes have exactly 3 divisors.
d) 1, 2, 4, 8, 16, 32
e) 1, 7, 25, 27, 16
7. a) Let’s take numbers 𝑎, 𝑏, and 𝑐 so that 𝑎|𝑏 and 𝑏|𝑐. To prove 𝑎|𝑐:
We know 𝑐 ÷ 𝑏 and 𝑏 ÷ 𝑎 are natural numbers and so 𝑐 ÷ 𝑎 = (𝑐 ÷ 𝑏) × (𝑏 ÷ 𝑎) is also a
natural number. Hence 𝑎|𝑐.
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