This is tough

i.mehrzad

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Feb 20, 2006
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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.
 
Gene said:
zero is not a positive integer.
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.
 
Whoops, foggy thinking. "p" must be positive but "a" doesn't. :oops:
--------------
Gene
 
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