Numbers and its Factors !!

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First express the given number 'N' in the form of f1^a x f2^b x f3^c.. where f1,f2,f3...fn are the prime factors of the given number 'N and a,b,c..are positive integers. Then the *Total no. of factors of 'N' is (a+1) x (b+1) x (c+1)*.... ...
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http://www.pagalguy.com/forums/quantitative-ability-and-di/number-system-questions-and-discussions-t-43728

http://www.pagalguy.com/forums/quantitative-ability-and-di/mind-blowing-concepts-at-quant-t-10313

www.raghavabbhi.com | My take on CAT: http://www.pagalguy.com/discussions/all-i-wanted-to-speak-about-cat-25002933/6315307
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First express the given number 'N' in the form of f1^a x f2^b x f3^c.. where f1,f2,f3...fn are the prime factors of the given number 'N and a,b,c..are positive integers.


Then the Total no. of factors of 'N' is (a+1) x (b+1) x (c+1)....

The sum of all factors (including 1 and the 'N' itself) is given by

[(f1^(a+1))/(f1-1)] x [(f2^(b+1))/(f2-1)] x [(f3^(c+1))/(f3-1)]....

The different ways of expressing the 'N with two different factors can be done in

[(a+1) x (b+1) x (c+1) ... -1 ] / 2 ways.

The different ways of expressing the 'N with two factors (including N^1/2 x N^1/2 ) can be done in

[(a+1) x (b+1) x (c+1) ... +1 ] / 2 ways.

The different ways of expressing the 'N with two co-primes can be done in

2^(A-1) ways.

Where A is the number of distinct prime factors of 'N'.

The number of co-prime of N that are less than N is given by :

N(1-(1/f1)) x (1-(1/f2))...

The sum of co-primes of N less than N is given by :

(N^2 /2) x (1-(1/f1)) x (1-(1/f2))... which can be simplified as (N/2)
x ( No. of co-primes of N less than N)

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Let us see with an example :

Let N be 48

48 can be expressed as 2^4 x 3^1

Here f1=2 and f2=3 and also a=4 and b=1.

So, No of factors of 48 is (4+1) x (1+1) = 5 x 2 = 10.

Sum of all factors of 48 is ((2^(4+1) -1)/2-1) x ((3^(1+1)-1)/3-1)= ((2^5 -1)/1) x ((3^2 -1)/2) 1 = ((32-1)/1) x ((9-1)/2) = 31 x 4 = 124.

The different ways of expressing the 48 with two different factors can be done in
[ (4+1) x (1+1) -1 ]/2 = 9/2 = 5 ways ( Just add 1 if the numerator comes as an odd value and then divide by 2).

The different ways of expressing the 48 with two factors can be done in
[ (4+1) x (1+1) + 1 ]/2 = 11/2 = 6 ways ( Just add 1 if the numerator comes as an odd value and then divide by 2).

The different ways of expressing the 48 with two co-primes can be done in

For 48, there are two distinct prime factors and therefore A is 2 here for us. So

2^(2-1)= 2 ways

The number of co-prime of 48 that are less than N is

48 x ( 1-(1/2)) x (1-(1/3)) = 48 X 1/2 X 2/3 = 16

The sum of all co-primes less than 48 is :

(48/2) x 16 = 384

Read '^' as 'to the power of' and 'x' as 'multiplied with' !!

Hope I have made you to understood the relationship between numbers and its factors in a detailed manner and do revert me back for any clarifications/doubts/corrections.

All the best puys !!








Regards, Ravi "Be Positive"
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