# Find all integers whose sum of divisors is equal to 48.

• March 19th 2012, 02:03 AM
math2011
Find all integers whose sum of divisors is equal to 48.
Let $\sigma(n)$ be the sum of all divisors of $n$. See sum-of-divisors function $\sigma_1$ definition in https://en.wikipedia.org/wiki/Divisor_function.

Solve $\sigma(n) = 48$.

The answers are $33,35,47$.

What are the strategies for solving this equation?

I can see that since $\sigma(n) = \Pi^s_{k=1} \frac{p_k^{\alpha_k + 1} - 1}{p_k - 1} = 48$, but there are too many factorizations of $48 = 2^4 \times 3$ to try.
• March 19th 2012, 02:23 AM
princeps
Re: Find all integers whose sum of divisors is equal to 48.
Quote:

Originally Posted by math2011
Let $\sigma(n)$ be the sum of all divisors of $n$. See sum-of-divisors function $\sigma_1$ definition in https://en.wikipedia.org/wiki/Divisor_function.

Solve $\sigma(n) = 48$.

The answers are $33,35,47$.

What are the strategies for solving this equation?

I can see that since $\sigma(n) = \Pi^s_{k=1} \frac{p_k^{\alpha_k + 1} - 1}{p_k - 1} = 48$, but there are too many factorizations of $48 = 2^4 \times 3$ to try.

If you are allowed to use computer then this simple Maple program solves your problem :

Code:

with(numtheory):
for n from 1 to 200 do
if sigma(n) = 48 then
print(n);
end if;
end do;

• March 19th 2012, 02:49 AM
math2011
Re: Find all integers whose sum of divisors is equal to 48.
Thanks, however I can't use Maple for this question.
• March 19th 2012, 05:48 AM
math2011
Re: Find all integers whose sum of divisors is equal to 48.
I found the method in my lecture notes. I will type post the solution tomorrow.
• March 20th 2012, 02:52 AM
math2011
Re: Find all integers whose sum of divisors is equal to 48.
The method in my lecture notes is only suitable for small numbers on the RHS of the equation.

Write down the values of $\sigma (p^{\alpha})$ for primes $p$ and integers $\alpha = 1,2,\ldots$, stop when the the value is greater than $\frac{48}{2}$ meaning it cannot be a factor or $48$.

$\sigma (2^\alpha)$: 3,7,15,31,...
$\sigma (3^\alpha)$: 4,13,40,...
$\sigma (5^\alpha)$: 6,31,156,...
$\sigma (7^\alpha)$: 8,57,...
$\sigma (11^\alpha)$: 12,133,...
$\sigma (13^\alpha)$: 14,183,...
$\sigma (17^\alpha)$: 18,...
$\sigma (19^\alpha)$: 20,...
$\sigma (23^\alpha)$: 24,...
$\vdots$:
$\sigma (47^\alpha)$: 48,...

Identify the ones that are factors of $48$.

$\sigma(3^1)=4$, $\sigma(5^1) = 6$, $\sigma(7^1) = 8$, $\sigma(11^1) = 12$, $\sigma(23^1) = 24$, $\sigma(47^1) = 48$.

Then

$\sigma(3^1) \sigma(11^1) = 48$ and so $3^1 \times 11^1 = 33$;
$\sigma(5^1) \sigma(7^1) = 48$ and so $5^1 \times 7^1 = 35$;
$\sigma(47^1) = 48$ and so $47^1 = 47$.