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.
Rate, Like and Share the posts here and help maximum people learn from it.
Cheers!
AS
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.
Rate, Like and Share the posts here and help maximum people learn from it.
Cheers!
AS
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