Find the sum of all proper divisors of:
N=2^p.3^q.5^r. wher p,q,r are positive integers.
Also find the sum of all prope divisors that are multiples of three.
Find the sum of all proper divisors of \(\displaystyle \L
N = 2^p \cdot 3^q \cdot 5^r\) is \(\displaystyle \L
\left( {\sum\limits_{a = 0}^p {\sum\limits_{b = 0}^q {\sum\limits_{c = 0}^r {2^a \cdot 3^b \cdot 5^c } } } } \right) - 2^p \cdot 3^q \cdot 5^r - 1\).
For the sum of all proper divisors that are multiples of three, use the same sum except change the starting index of b from 0 to 1 and do not subtract the 1.
I agree, except that as I recall, zero is not a positive integer. I would change the indices to a=1 etc. That drops the -1 and they are all multiples of 3.
That is absolutely correct, zero is not a positive integer.
But then the sum does not imply that it is!
If \(\displaystyle \L
a = 0\; \wedge b = 0\; \wedge c = 0\quad \Rightarrow \quad 2^0 \cdot 3^0 \cdot 5^0 = 1\) and 1 is divisor albeit not a proper divisor.
You see we must begin all the indices at 0 in order to get the divisors 2, 3 and 5.
Beginning the ‘b’ sum at 1 forces multiples of 3 and improper divisor 1 is thus eliminated.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.