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How to find total no of divisors of any number and how to calculate sum of all factors of any natural number?Please explain it with an example.
Answers (1)
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This concept is given in our study material. Please remember the formula for such question.
If your number n = a^x * b^y * c^z (where a, b, and c are n's prime divisors and x, y, and z are the number of times that divisor is repeated) then the total count for all of the divisors is (x + 1) * (y + 1) * (z + 1)Let N be a natural number
such that N= ap . bq . cr
therefore the sum of the Factors of N = (ap+1 - 1)(bq+1 -1)(cr+1 - 1)/(a-1)(b-1)(c-1)