Factors - Fundamentals 1

This article discusses various fundamental properties associated with factors of a number. I will present some pointers with a solved example and provide a practice question with the answer to help you learn as you read and feel comfortable with the topic at the end of the article.

Let's unveil what lies behind the curtains of this drama of factors..

POINT 1)
N = a^p * b^q * c^r * ...
where a,b,c,.. are DISTINCT PRIMES.
then,
Total number of factors(f) = (p + 1)(q + 1)(r + 1)..

Also,if N = 2^p * b^q * c^r * ... form
then,
Number of even factors = p*(q + 1)(r + 1)...
Number of odd factors = (q + 1)(r + 1).. i.e. ignore the power of 2 and calculate for the rest.
example : N = 20
20 = 2^2 * 5^1
Total number of factors = (2 + 1)(1 + 1) = 6
Even factors = 2*(1 + 1) = 4
Odd factors = (1 + 1) = 2 = Total - Even

Practice Question : Find the number of odd,even and total factors for :
i) N = 360
ii) N = 105
Answer : i) Total : 24 , Even : 18 , Odd = 6  ; ii) Total : 8, Even : 0 , Odd : 8


POINT 2
N = a^p * b^q * c^r * ... where a,b,c are DISTINCT PRIMES
then,
Sum of all the factors = {[a^(p + 1) - 1][b^(q + 1) - 1]...}/(a - 1)(b - 1)..
This can also be written as : (1 + a + a^2 + .. + a^p)(1 + b + b^2 + ..b^q)..
NOTE : Sum of powers of prime starting from 1
2^0 + 2^1 + 2^2 + 2^3 = (2^4 - 1)/(2 - 1) = 15
3^0 + 3^1 + 3^2 = (3^3 - 1)/(3 - 1) = 13

example : N = 10 = 2^1 * 5^1
Sum of all the factors = [2^(1 + 1) - 1][5^(1 + 1) - 1] / [(2 - 1)(5 - 1)
= (4 - 1)(25 - 1)/(1)(4) = 18
Factors of 10 = {1,2,5,10} = Sum = 18 = Verified.

The second form of the sum of factors of N is a product of sum of power of primes from 0 to n that occur in its prime factorization.
Hence if sum = S is given. We try to write it as product of sums of powers of primes and match it with standard form to find the N.

Now For a particular N, S = constant.
But for a particular S, N may or may not be unique or may not even exist.
For this you need practice and acquaintance with
1 + 2 + 4 = 7 = 2^3 - 1
1 + 5 + 25 = 31
1 + 7 = 8 while 1 + 7 + 49 = 57
You need to work with them multiples times and their product combinations to be familiar with this.

example : If sum of all factors of N is 18. 18 = 3*6 = (2^0 + 2^1)(5^0 + 5^1)
=> N = 2^1 * 5^1 = 10. No other factorization of 18 would give such a form
so N = 10 only.

Practice Question :
i) Find sum of all factors of 120
ii) If sum of all the factors of N is 31. Find N.
Answer :
120 = 2^3 * 3 * 5
Sum of factors = (16 - 1)(9 - 1)(25 - 1)/(2 - 1)(3 - 1)(5 - 1) = 360

Sum of factors = 31 = (2^0 + 2^1 + 2^2 + 2^3 + 2^4) OR (5^0 + 5^1 + 5^2)
Hence N = 2^4 or 5^2 = 16 or 25


POINT 3)N = a^p * b^q * c^r * ... where a,b,c are DISTINCT PRIMES
then,
i) Number of factors of N that are perfect squares
( [p/2] + 1)( [q/2] + 1)( [r/2] + 1)... where [.] is Greatest Integer Function
which gives the greatest integer less than or equal to value inside it.
[2.5] = 2 ; [integer] = integer itself ; [-2.5] = -3

ii) Number of factors of N that are perfect cubes
( [p/3] + 1)( [q/3] + 1)( [r/3] + 1) ...

The Logic behind the formula :
For perfect squares we have 2^0,2^1,...,2^10 available(similarly for other powers) and to keep a factor perfect square we can use :
2^0 or 2^2 or 2^4 .. or 2^10 which is basically 0 to 10 in steps of 2
So [10/2] + 1 = 6 = { 0,2,4,6,8,10 }
Similarly for perfect cubes.

example :
N = 2^2 * 3^3 * 4^4 * 5^5
N = 2^10 * 3^3 * 5^5
Factors that are perfect square
= ( [10/2] + 1)( [3/2] + 1)( [5/2] + 1)
= ( 5 + 1 )( 1 + 1) ( 2 + 1) = 6 * 2 * 3 = 36
Factors that are perfect cube
= ( [10/3] + 1)( [3/3] + 1)( [5/3] + 1)
= (3 + 1)( 1 + 1)( 1 + 1) = 4*2*2 = 16

This should be enough for a single-go read and I will follow this up with a Factors - Fundamentals 2 to discuss the remaining pointers.

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Cheers!
AS