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

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

Solve $\displaystyle \sigma(n) = 48$.

The answers are $\displaystyle 33,35,47$.

What are the strategies for solving this equation?

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

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

Quote:

Originally Posted by

**math2011** Let $\displaystyle \sigma(n)$ be the sum of all divisors of $\displaystyle n$. See

**sum-of-divisors function **$\displaystyle \sigma_1$ definition in

https://en.wikipedia.org/wiki/Divisor_function.

Solve $\displaystyle \sigma(n) = 48$.

The answers are $\displaystyle 33,35,47$.

What are the strategies for solving this equation?

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

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

Thanks, however I can't use Maple for this question.

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.

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

$\displaystyle \sigma (2^\alpha)$: 3,7,15,31,...

$\displaystyle \sigma (3^\alpha)$: 4,13,40,...

$\displaystyle \sigma (5^\alpha)$: 6,31,156,...

$\displaystyle \sigma (7^\alpha)$: 8,57,...

$\displaystyle \sigma (11^\alpha)$: 12,133,...

$\displaystyle \sigma (13^\alpha)$: 14,183,...

$\displaystyle \sigma (17^\alpha)$: 18,...

$\displaystyle \sigma (19^\alpha)$: 20,...

$\displaystyle \sigma (23^\alpha)$: 24,...

$\displaystyle \vdots$:

$\displaystyle \sigma (47^\alpha)$: 48,...

Identify the ones that are factors of $\displaystyle 48$.

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

Then

$\displaystyle \sigma(3^1) \sigma(11^1) = 48$ and so $\displaystyle 3^1 \times 11^1 = 33$;

$\displaystyle \sigma(5^1) \sigma(7^1) = 48$ and so $\displaystyle 5^1 \times 7^1 = 35$;

$\displaystyle \sigma(47^1) = 48 $ and so $\displaystyle 47^1 = 47$.