Given that 8191 is a Mersenne prime, find the last three digits of the even perfect number N that has 8191 as a factor, given that N is less than 1 billion.
Details and assumptions
A perfect number is equal to the sum of its proper divisors (divisors excluding itself).
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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 8 1 9 1 + 1 = 8 1 9 2 is a power of 2. lo g 2 8 1 9 2 = 1 3 , i.e. 8192 is the 13th power of 2, and so p = 1 3 .
The corresponding perfect number is: 2 1 2 ∗ ( 2 1 3 − 1 ) = 3 3 , 5 5 0 , 3 3 6 , which is less than a billion as was requested, and so the answer is: 3 3 6
Latex mistakes!! This is what they were supposed to be:
the number 2 p − 1 ( 2 p − 1 ) is an even perfect number.
The corresponding perfect number is: 2 1 2 ( 2 1 3 − 1 ) = 3 3 , 5 5 0 , 3 3 6
I need to get better at this Latex thing... and an edit option for solutions would be wonderful.
If 2 p − 1 is prime, then 2 p − 1 ( 2 p − 1 ) is a perfect number.
⟹ 8 1 9 1 = 2 1 3 − 1
So, it is associated with the perfect number 2 1 3 − 1 ( 2 1 3 − 1 ) = 3 3 5 5 0 3 3 6
Clearly it has 8191 as a factor.
The Last three digits of N = 3 3 6
Perfect numbers can be generated by the formula 2 p − 1 ⋅ ( 2 p − 1 ) . Here 2 p − 1 is a Mersenne prime. The question reveals the value of 2 p − 1 = 8 1 9 1 = 8 1 9 2 − 1 = 2 1 3 − 1 . Hence, p = 1 3 , the required last three digits are... 2 1 3 − 1 ⋅ ( 2 1 3 − 1 ) = 2 1 2 ⋅ 8 1 9 1 = 4 0 9 6 ⋅ 8 1 9 1 ≡ 9 6 ⋅ 1 9 1 = 1 8 3 3 6 ≡ 3 3 6 ( m o d 1 0 0 0 )
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No code required...Only a little computation required to know that 8 1 9 1 = 2 1 3 − 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 1 3 − 1 is a prime ( 2 1 2 ) ( 2 1 3 − 1 ) must be an even perfect number. This number is 33550336 giving us the answer 336