Division, remainders and prime numbers Addition and multiplication ...

Division, remainders and prime numbers
Addition and multiplication laws for remainders
1) The remainder of a sum of two numbers is given by the sum of their remainders.
2) The remainder of the product of two numbers is given by the product of their
remainders.
For example consider the remainders of division by 3:
There are three possibilities: The remainder can be 0, 1 or 2.
Now observe:
The sum 15 = 5 + 10 has remainder given by the sum of remainders 2 + 1 = 3 so the
remainder is 0. (daaah!!)
50 = 5 · 10 has remainder given by 2 · 1 = 2.
8 · 14 has remainder given by 2 · 2 = 4, so the remainder is 1.
(We could also allow negative remainders:
We can say that the remainder of division of 6 by 7 is −1.
The remainder of division of 16 by 7 is 2 or −5.
Same rules apply.)
Why do the laws work?
Main easy point: if two numbers differ by a multiple of 3 they leave the same remainder
when divided by 3.
(For example the numbers 17 and 8 differ by 9 = multiple of 3. So they leave the
same remainder 2.)
Now say we want to divide a and b by 3: we write
a = 3 · q + r, q =quotient, r =remainder = 0, 1, 2
b = 3 · q ′ + r ′ , q ′ =quotient, r ′ =remainder = 0, 1, 2
Now add up:
a + b = 3 · q + r + 3 · q ′ + r ′ = 3 · q + 3 · q ′ + (r + r ′ )
So a + b differs from r + r ′ by a multiple of 3; this does not change the remainder.
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Problem 1: What is the remainder of division of 8 2012 by 7?
Problem 2: What is the remainder of division of 2 2012 by 3? how about of 2 2011 ?
Problem 3: What are the remainders of division of 10, 100, 1, 000, 1, 000, 000 by 3?
Problem 4: What is the remainder of division of 40, 021 by 3?
(Hint: Write 40, 021 = 4 · 10, 000 + 21.)
Problem 5: What is the remainder of division of 2 2012 by 7?
Problem 6: What is the remainder of division of 10, 100, 1000 by 11?
Problem 7: What is the remainder of division of 35, 221 by 11?
Problem 8: Explain why the fourth power a 4 of any number a, when divided by 5,
gives remainder only either 0 or 1. What happens with the fifth power a 5 ?
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Prime numbers: Their only factors are themselves and 1.
(the number 1 is not considered a prime number)
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, . . .
(Obvious) fact: Every number larger than 1 is a product of prime numbers.
Examples:
168 = 2 · 2 · 2 · 3 · 7,
238 = 2 · 7 · 17,
1, 024 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 = 2 10 .
Problem 1:
a) Is 3 a factor of 234?
b) Is 3 a factor of 45, 673? of 5, 682, 985?
c) Is 11 a factor of 275, 385?
Problem 2: Write the following numbers as a product of prime numbers:
a) 420=
b) 408=
c) 2,012=
d) 13506641086599522334960321627880596993888147560566702752448514385152651060
48595338339402871505719094417982072821644715513736804197039641917430464965
89274256239341020864383202110372958725762358509643110564073501508187510676
59462920556368552947521350085287941637732853390610975054433499981115005697
7236890927563=
(Has 309 digits: A cash price of $100,000 was offered for finding one factor of this
number!!!! Hint that does not help: this number is a product of just two primes.)
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(FOR CHEATERS: go to http://www.alpertron.com.ar/ECM.HTM )
Problem 3: Are the following numbers prime or not?
a) 163
b) 247
c) 11,111,111
d) 691
Problem 4: Let’s say x is a number between 1, 000 and 2, 000. We check that x is not
divisible by any prime number between 2 and 43 (these two included). Is the number x
prime?
Problem 5: Consider the number (2 · 3 · 5 · 7 · 11 · 13 · 17 · 19) + 1 = 9, 699, 691.
Is it divisible by a number between 2 and 20?
Problem 6: Find two numbers a and b (could be negative) such that
1 = 13 · a + 17 · b.
Problem 7: The same problem as above but for any two different primes p and q
instead of 13 and 17. Is there a method that always work to find a and b such that
1 = p · a + q · b ?
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How many primes are there?
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, . . . . . .
Suppose that the list ends and that there is a largest prime=godzillion, blaa, ......
I can then take the product of all the primes in the list
2 · 3 · 5 · 7 · 11 · · · · · · (largest prime)
and I add one to it:
2 · 3 · 5 · 7 · 11 · · · · · · (largest prime) + 1.
This (huge) number is not divisible by 2, 3, . . .. It is not divisible by any prime in
our list so far. It has, however, a prime factor for sure. This prime factor is a prime
number which is bigger than all the prime numbers in our list.
So the list never ends; the list of primes is infinite!!
Question 1: What is the largest prime which is known today?
Answer: 2 43,112,609 − 1 (has more than 10 million digits)
Question 2: What is the “probability” that a number is prime?
how many primes are less than 10= 4, 40%
how many primes are less than 100=25, 25%
how many primes are less than 1,000=168, 16,8%
how many primes are less than 10,000=1,229, 12,3%
how many primes are less than 100,000=9,592, 9,6%
how many primes are less than 1,000,000=78,498, 7,85%
Answer: Roughly decreases proportionally to the number of digits we allow the
number to have.
Question 3: Is every even number (except 2) a sum of two primes?
(we are allowed to use the same prime twice, for example: 6 = 3 + 3.)
4 = 2 + 2, 6 = 3 + 3, 8 = 3 + 5
10 = 7 + 3 = 5 + 5, 12 = , 14 = , 16 = ,
18 = , 20 = , 22 = , 24 = , 100 =
Question 4: 3 and 5 are twin primes (they are as close as possible), same for 5 and
7, 11 and 13, and also 29 and 31. Is the list of such pairs of twin primes infinite?
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