8191 Perfect Numbers

No code required...Only a little computation required to know that $8191=2^{13}-1$ since it is a Mersenne prime. From Euclid's proposition:

$(2^{p-1})(2^{p}-1)$ * is an even perfect number whenever $2^{p}-1$ is prime . *

So since $2^{13}-1$ is a prime $(2^{12})(2^{13}-1)$ must be an even perfect number. This number is 33550336 giving us the answer 336

There is a very interesting correlation between Mersenne primes and perfect numbers...

It is known that for any Mersenne prime, of the form $2^p - 1$ , the number $2^(p-1) (2^p - 1)$ is an even perfect number. 8191 is a Mersenne prime, so $8191 + 1 = 8192$ is a power of 2. $\log_2{8192} = 13$ , i.e. 8192 is the 13th power of 2, and so $p = 13$ .

The corresponding perfect number is: $2^12 * (2^13 - 1) = 33,550,336$ , which is less than a billion as was requested, and so the answer is: $\fbox{336}$

If $2^p-1$ is prime, then $2^p-1(2^p-1)$ is a perfect number.

$\implies 8191=2^{13}-1$

So, it is associated with the perfect number $2^{13-1}(2^{13}-1)= 33550336$

Clearly it has 8191 as a factor.

The Last three digits of ${N}=\boxed{336}$

Perfect numbers can be generated by the formula $2^{p-1} \cdot (2^p-1)$ . Here $2^p-1$ is a Mersenne prime. The question reveals the value of $2^p-1=8191=8192-1=2^{13}-1$ . Hence, $p=13$ , the required last three digits are... $2^{13-1}\cdot (2^{13}-1)=2^{12}\cdot 8191 =4096\cdot 8191\equiv 96 \cdot 191 = 18336 \equiv \fbox{336}\pmod{1000}$