sum of all even and odd positive divisors

Question

Find the sum of all odd divisors of 3600 $3600=6^2.10^2$ Divisors are of the form $6^α.10^β$ Each term in the product $(1+6+6^2)(1+10+10^2)$ is a positive divisor of $3600$ Now how do I separate even and odd divisors? Any idea is appreciated

$6$ and $10$ are not primes. $2$, $3$ and $5$, for example, are divisors not of that form. — Robert Israel, Jun 14 '18 at 19:01

Rhys Hughes · Accepted Answer · 2018-06-15T15:59:09.560

1

$$3600=2^4\cdot3^2\cdot5^2$$ Hence, any positive divisors of $3600$ are of the form $$2^a\cdot3^b\cdot5^c \text{ with }0\le a\le4, 0\le b\le2, 0\le c\le2; a,b,c\in \Bbb Z$$

The odd divisors are those with $a=0$*, and the even divisors are those with $a\ne0$.

*This is because $2^0=1$, which means your number is a multiple of only odd numbers and is therefore odd.

Hence, for your odd divisors, they are of the form $3^b\cdot5^c$, with $b,c\in(0,1,2)$. Vary $b$ and $c$ between those three values and add up all $9$ results.

edited Jun 15 '18 at 15:59

answered Jun 14 '18 at 19:15

Rhys Hughes

12,842

$2$ more lines please – user568963 Jun 15 '18 at 08:07
Added $2$ more lines. Hope it helps – Rhys Hughes Jun 15 '18 at 15:59
Thanks it helped – user568963 Jun 17 '18 at 06:24

sum of all even and odd positive divisors

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