workout wrote:
Bunuel wrote:
NUMBER THEORY
Finding the Sum of the Factors of an Integer
First make prime factorization of an integer\(n=a^p*b^q*c^r\) , where\(a\) ,\(b\) , and\(c\) are prime factors of\(n\) and\(p\) ,\(q\) , and\(r\) are their powers.
The sum of factors of\(n\) will be expressed by the formula:\(\frac{(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)}{(a-1)*(b-1)*(c-1)}\)
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Finding the Sum of the Factors of an Integer
First make prime factorization of an integer\(n=a^p*b^q*c^r\) , where\(a\) ,\(b\) , and\(c\) are prime factors of\(n\) and\(p\) ,\(q\) , and\(r\) are their powers.
The sum of factors of\(n\) will be expressed by the formula:\(\frac{(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)}{(a-1)*(b-1)*(c-1)}\)
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