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logic behind finding the number of factors

Asked by ashwinivg 27th April 2010, 3:17 PM

Dear student,

The logic is as follows -

Lets say we need to find out the factors of a number N.

We prime factorize it in the form say 2k.

(for the sake of simplicity, I have taken only one prime factor here. We can easily extend the concept to more than one prime factors.)

Now, any number of the form 2i (where i is an integer such that 0<= i <= k) is also a factor of N, as it divides 2k which in turn divides N.

So for the values of i in set S = {0, 1, 2 , 3, 4, .... , k}, we have a factor of N. ( for i=0, 2i = 1, which is always a factor of every real number).

Since there are (k + 1) values in the set S, therefore the number of factors for N are k+1.

For your example, the value of k = 5.

i=0 , 20=1 is a factor.

i=1 , 21=2 is a factor.

i=2 , 22=4 is a factor.

i=3 , 23=8 is a factor.

i=4 , 24=16 is a factor.

i=5 , 25=32 is a factor.

=> 5+1 = 6 factors.

Answered by Expert 27th April 2010, 9:11 PM
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