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AS

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Factors - Fundamentals 3

Follow up to Factors - Fundamentals 1 and 2
Ordered and Unordered Sets, in simple words, can be seen as follows :

Ordered sets :
If for a given situation,a set of n elements satisfies the problem solution then all permutations of those n elements that satisfy the problem solution are to be counted as different i.e

a,b is treated different than b,a ; a,b,c is counted different from a,c,b or b,a,c or b,c,a or c,a,b or c,b,a

Unordered sets :
If for a given situation,a set of n elements satisfies the problem solution then any permutation of that set satisfying the problem solution should be considered as the same and not counted again.

a,b is same as b,a and a,b,c is same as other (3! - 1) = 5 permutations.

POINT 1)


Number N as a product of its two factors.
f : Total number of factors of N.
N = a*b
N is a perfect square :
If N is a perfect square it contains one solution as N = a*a = a^2 and rest are (a,b) distinct format.
Hence,
Ordered(distinct) = (f - 1)
Ordered(not necessarily distinct) = f
Unordered(distinct) = (f - 1)/2
Unordered(not necessarily distinct) = (f - 1)/2 + 1 OR (f + 1)/2

N is not a perfect square
Ordered(any case) = f
Unordered(any case) = f/2

example :
N = 100 = 2^2 * 5^2
f = 3*3 = 9

100 = a*b
Then if a is some factor of 100,then b = 100/(that factor) and vice versa.
Hence we have as many (a,b) sets satisfying the equation as are the number of factors of N.
But since N = 100 = perfect square hence it has one solution : 10,10 counted twice.
1*100,100*1,2*50,50*2,...10(first)*10(second),10(second)*10(first)
So,
Ordered(distinct) = f - 1 = 8
Ordered(not necessarily distinct) = f = 9
Unordered(distinct) = (f - 1)/2 = 4
Unordered(not necessarily distinct) = (f - 1)/2 + 1 OR (f + 1)/2 = 5


POINT 2)

Number of ways to resolve N into product of two co-primes = 2^(n - 1) where n is the number of distinct primes in the prime factorization of N.
example :
N = 12 = 2^2 * 3
Then, answer = 2^(2 - 1) = 2 ; As there are two primes : 2 and 3 in factorization of N
We can see that as :
1*12 : Co-prime
2*6 : HCF(2,6) = 2
3*4 : Co-prime
2 sets.


POINT 3)

Number of ways to resolve N into product of three co primes.

example :
N = 2^3 * 3^2 * 5^4
N = abc such that HCF(a,b,c) = 1 i.e they are co-prime.

a = 2^x1 * 3^y1 * 5^z1
b = 2^x2 * 3^y2 * 5^z2
c = 2^x3 * 3^y3 * 5^z3

x1 + y1 + z1 = 3 => 5C2 ways. Out of this when none of them is zero needs to be removed as in that case
HCF would not be 1.
So, 5C2 - 1 = 9 ways
Similarly,
x2 + y2 + z2 = 2 => 4C2 = 6 ways. With none of them zero the equation cannot hold for non negative integers.
Hence nothing is to be subtracted.
Also,
x3 + y3 + z3 = 4 => 6C2 - 3 = 12 ways

Ordered = 9*6*12 = 648
Unordered = (648 - 3!/2!*4)/3! + 4 = 110
[1,1,n ; 3,3,x ; 25,25,y ; 75,75,z ] being the 4 triplets to be removed for unordering ]

That brings us to close of the 3 part series of Factors - Fundamentals article. These 3 part should allow you to solve most of the problems related to Factors of a number barring a few typical concepts which will be discussed as separate special article devoted to that topic alone later down the line. Till then, keep following the blog for such illustrative concepts.

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Cheers!
AS

0

Factors - Fundamentals 2

This is a continued post of the Factors - Fundamentals 1 and will dive into the remaining pointers associated with the problems based on factors of a number. We have earlier discussed the number of factors, sum of factors, find number when sum of factors given and also the number of perfect square or cube factors of any number. Let's follow them up with some more important pointers today beginning with ways to write a number as sum of consecutive numbers..

POINT 1

i) Number of ways in which N can we written as a sum of consecutive natural numbers is equal to
(No of odd factors - 1)
Logic :
Let the k consecutive numbers be a,a+1,a+2,..
N = Sum = k/2 * [a + a + (k - 1)]
= k/2 * [ 2a + (k - 1) = k * [ a + (k - 1)/2 ]
So k is a factor of N and (k - 1) must be even for N to be an integer => k must be odd => k must be an ODD factor of N
But if k = 1 then we can't call it sum of "consecutive"
So No of odd factors of N - 1

ii) Number of ways in which N can we written as a sum of consecutive INTEGERS = 2*(Number of odd factors) - 1

example : Find the number of ways 105 can be written as sum of consecutive
i) Natural numbers
ii) Integers.105 = 3*5*7
Number of odd factors = 2*2*2 = 8
So answer for part i) = 8 - 1 = 7 and part ii) 2*8 - 1 = 15

What's really happening here is.
Take one of the odd factors ,say, 5
Now, 5*21 = 105
So what we do is we take one number as 21 and split remaining 4 on either side of 21 as 19,20 , [21] , 22 , 23
And these form the sum = 105 and stay consecutive.


POINT 2i) Product of all factors of N is given by : N^(f/2) where f is the number of factors of N

example : N = 10
Factors = 1,2,5,10
So these factors form N in pairs from extreme ends :
1*10 = 10 ; 2*5 = 10
So product of all factors = 1*2*5*10 = (1*10)*(2*5) = 10^2 = 10^(4/2)
Always like this.

ii) Sum of all numbers less than N and co prime to it is given by : N * E(N)/2 where E(N) is the Euler number.
example : N = 2^2 * 3^3 * 5^5

E(N) = N*(1 - 1/a)(1 - 1/b)(1 - 1/c)... where a,b,c are the DISTINCT primes in prime factorization of N

So in our case E(N) = 2^2 * 3^3 * 5^5 * (1 - 1/2)(1 - 1/3)(1 - 1/5)
= 2^2 * 3^3 * 5^5 * 1/2 * 2/3 * 4/5
= 2^4 * 3^2 * 5^4 = 16*9*625 = whatever it is.
So sum of all numbers less than N and co prime to N = N*E(N)/2
Substitute the values.

POINT 3
i) Perfect squares have odd number of factors.

N = perfect square = a^even * b^even * c^even form always where a,b,c are distinct primes
Number of factors = (even + 1)*(even + 1)*(even + 1) = odd*odd*odd = odd

(*) Only squares of prime have exactly 3 factors.[Because they are just a^2 form where a is the prime]

ii) Factors of N^2 less than N but not a factor of N is given as :
(Factors of N^2 - 1)/2 - (Factors of N) + 1

One of the factors of N^2 is N.
By extension all factors of N are factors of N^2 too.
We need those factors of N^2 that are less than N yet not a factor of N
By symmetry half of the factors of N^2 shall lie below and half above N about the vertex point N itself.
So first part is that division to reach below N part. Then second part removes Factors of N from it.When in first part we divide about N and subtract factors of N we subtract N once[Factors of N includes N itself as well] which was already not counted. Hence +1 for that in the end..

example : Find the number of factors of 144 that are less than 12 but not a factor of N12 = 2^2 * 3 => f(12) = 3*2 = 6
144= 2^4 * 3^2 => f(144) = 5*3 = 15
(15 - 1)/2 - 6 + 1 = 2


That concludes this segment on Factors of a number and its related fundamental properties. This should help you gain a hold on basic problems relating to factors that appear in the MBA examinations. Part 3 of this series to follow soon.

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Cheers!
AS

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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

0

Remainders Unlocked : Special cases

This is the final post on remainders wherein I will look to bring all the question specific concepts relating to remainders into this. So, let's start straight away..

Using Binomial Theorem :

(a + b)^n = C(n,0)*a^n + C(n,1)*a^(n-1)*b + C(n,2)*a^(n-2)*b^2 + .... C(n,n)*b^n

example : Find the remainder when 7^25 is divided by 36.
7^25 = (6 + 1)^25 = C(25,0)*6^25 + C(25,1)*6^24 + ... + C(25,23)*6^2 + C(25,24)*6 + C(25,25)
All terms except last two terms will be divisible by 36.
So, 7^25 mod 36 = [C(25,24)*6 + C(25,25)] mod 36 = (25*6 + 1) mod 36 = 151 mod 36 = 7.


Using factor theorem and remainder theorem :


If g(x) divides f(x), we say that f(x) is divisible by g(x) or g(x) is a factor of f(x).
Also, when a polynomial f(x) is divided by (x - a) then remainder is given by f(a)

example 1 : What is the remainder when x^4 - 3x^2 + 1 is divided by (x - 2)
Let f(x) = x^4 - 3x^2 + 1. Then, remainder = 2^4 - 3*2^2 + 1 = 5

example 2 : What is the largest value of n for which n^3 + 100 is divisible by (n  + 10).
We substitute n = -10 in the polynomial to get (-10)^3 + 100 = -900
(n + 10) must be now a factor of 900
k(n + 10) = 900
For largest value of n, k = 1
n = 890


Using Divisibility Rules :

Divisibility by 2 , 4(=2^2) , 8(=2^3) , 16(2^4) , ...

You check divisibility and remainder by last n digits of the number when number divided by 2^n.

Divisibility by 3

You check divisibility and remainder by calculating sum of digits of the number and dividing that by 3 or 9.

Divisibility by 5, 25(=5^2) , 125(=5^3) , ...

You check divisibility and remainder by last n digits of the number when divided by 5^n

Divisibility by numbers of the form 10^n + 1 : 11, 101, 1001 , ...

Break the number in groups of n digits starting from the right and add the n-digit groups with alternate + and - signs. If the sum is divisible by 10^n + 1, then the number is divisible by 10^n + 1

Why this happens?

Any number abcd(say) can be written as : a*10^3 + b*10^2 + c*10 + d
Now let us take remainder by 11.
Since 10 mod 11 = 10 or -1. 
We have 10^n mod 11 = -1 when n is odd and 10^n mod 11 = 1 when n is even
Hence we get, abcd mod 11 = (-a + b - c + d) mod 11

Divisibility by numbers of the form 10^n - 1 : 9, 99, 999, ..


Break the number in groups of n digits starting from the right and add the n-digit groups. If the sum is divisible by 10^n - 1, then the number is divisible by 10^n - 1

Why this happens?
 
Since 10 mod 9 = 100 mod 99 = 1000 mod 999 = 1 and so on,
So writing a number say abcd as ab*10^2 + cd, we can find remainder by 99 as :
(ab*10^2 + cd) mod 99 = (ab + cd) mod 99 which is nothing but groups of 2 from the right.


Divisibility by any prime greater than 5 : Seed number method

How to find seed number ?

We need to find a multiple of prime number that ends up in 10k +/- 1 form. Usually this would happen within the first 10 multiples of the prime.
example : Let 13 be the prime.
13*3 = 39 = 4*10 - 1 form => Seed number = 4
13*7 = 91 = 9*10 + 1 form => Seed number = (-9) 

NOTE : Seed number method is applicable to other base systems too with the only change that you look for bk +/- 1 format for a given base b

How to use seed number for divisibility ?

Break the number into unit digit,denoted by B and remaining digits,denoted by A.
Now if the seed number is N, then check for divisibility for A + NB and if the seed number is (-N) then check for divisibility for A - NB.


This brings us to close of the topic of remainders with the only discussion left in context of cyclicity and base systems other than decimal(base 10). These topics are independently big enough so will discuss them in separate articles some other day.

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Cheers!
AS

0

Remainders Unlocked : Patterns and Generalizations

This is the second post on remainders. Earlier one focused on standard important theorems that are used in remainder problems. Now we will be delving into details of some other important rules,generalizations,patterns,etc. I'll discuss these as pointers and provide examples along the way if it seems suitable. Let's start then :)

Point 1 : (a^n + b^n) mod (a + b) = 0 when n is odd
(a^n - b^n) mod (a + b) = 0 when n is even

(a^n - b^n) mod (a - b) = 0 for all natural numbers

(a^n - b^n) mod (a^k - b^k) = 0 when k is a factor of n

example : 18^2000 + 12^2000 - 5^2000 - 1 is divisible by ?
323,221,299,237

(18^2000 - 5^2000) + (12^2000 - 1^2000) = B1 + B2

B1 is divisible by 13 by result 3 and B2 is divisible by 13 by result 2. Hence the expression is divisible by 13.
Also,writing it as :
(18^2000 - 1^2000) + (12^2000 - 5^2000) = B1 + B2

B1 is divisible by 17 by result 3 and B2 is divisible by 17 by result 2. Hence the expression is divisible by 17.
Hence the expression is divisible by 13*17 = 221

Point 2 :

Any number of the abcabc form is divisible by 1001 and hence by extension by 7,11 and 13 as
1001 = 7*11*13

Point 3 :

If 10^n mod x = 1, then to find remainder for any number upon division by x, you can take n digits at a time, find individual remainders and simply add them up.

example : Find the remainder when 123123123123123...upto 300 digits is divided by 27.

Since 999 = 27*37. So, 1000 mod 27 = 1
Hence we can take three digits at a time and find individual remainders.
123 mod 27 = 15 and for 300 digits we'll have 300/3 = 100 such sets.
So, finally we have 15*100 mod 27 = 15 = Remainder.

Point 4 :

(a^n + b^n + c^n + ...) mod (a + b + c + ...) = 0 if and only if a,b,c,.. are in AP and n is ODD

example : 16^3 + 17^3 + 18^3 + 19^3 mod 70 = ?

Answer : Zero. As 16,17,18,19 are in AP and 3 is odd. Hence the expression is divisible by sum of the base terms : (16 + 17 + 18 + 19) = 70.

Point 5 : 

Any digit or a set of digits repeated (p - 1) times is divisible by p.

example : 121212121212 mod 7 = ?

Answer : Zero. Since 12 12 12 12 12 12 is a repeated set of 12 coming 6(=7 - 1) times. Hence is divisible by 7.

NOTE : The topic of remainders becomes problematic as there is no end to the generalizations that one can make out of various set of questions. So you always end up finding a new concept or pattern or a generalization. Best way is to keep in mind these important and common ones and add up plus make note of what you keep finding by practicing more and more.

These cover majority of remainder questions. Other techniques are specific to question type at hand or involve different concepts like cyclicity,divisibility rules or base system for example. We will see them someday later when we come to those discussions individually.

Till then, hope you enjoyed learning from my blog and share it wherever you can. Rate, like and share the posts through the social network links at the end of each post. Leave comments for any doubts or personal reviews of how you felt about the article.

Cheers!

AS 

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