8191 Perfect Numbers

Level pending

Given that 8191 is a Mersenne prime, find the last three digits of the even perfect number N N that has 8191 as a factor, given that N N is less than 1 billion.

Details and assumptions

A perfect number is equal to the sum of its proper divisors (divisors excluding itself).


The answer is 336.

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4 solutions

Thaddeus Abiy
Dec 15, 2013

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

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

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

This presupposes that N N is unique, which need not be the case. You need to account for the possibility of an odd perfect number, or at least mention that none is currently known (and they have checked past 1 billion).

Calvin Lin Staff - 7 years, 5 months ago
Ben Frankel
Dec 15, 2013

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

The corresponding perfect number is: 2 1 2 ( 2 1 3 1 ) = 33 , 550 , 336 2^12 * (2^13 - 1) = 33,550,336 , which is less than a billion as was requested, and so the answer is: 336 \fbox{336}

Latex mistakes!! This is what they were supposed to be:

the number 2 p 1 ( 2 p 1 ) 2^{p-1} (2^p - 1) is an even perfect number.

The corresponding perfect number is: 2 12 ( 2 13 1 ) = 33 , 550 , 336 2^{12} (2^{13} - 1) = 33,550,336

I need to get better at this Latex thing... and an edit option for solutions would be wonderful.

Ben Frankel - 7 years, 5 months ago
Manoj Pandey
Dec 17, 2013

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

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

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

Clearly it has 8191 as a factor.

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

Jubayer Nirjhor
Dec 15, 2013

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

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