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

**math2011** · March 19th 2012, 02:03 AM

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.

**princeps** · March 19th 2012, 02:23 AM

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;

**math2011** · March 19th 2012, 02:49 AM

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

**math2011** · March 19th 2012, 05:48 AM

I found the method in my lecture notes. I will type post the solution tomorrow.

**math2011** · March 20th 2012, 02:52 AM

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 $47^1 = 47$ .

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

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