Math & Science Home | Proficiency Tests | Mathematical Thinking in Physics | Aeronauts 2000 |
||||
Perfect Numbers - A Case StudyPerfect numbers are those numbers that equal the sum of all their divisors including 1 and excluding the number itself. Most numbers do not fit this description. At the heart of every perfect number is a Mersenne prime. All of the other divisors are either powers of 2 or powers of 2 times the Mersenne prime. Let's examine the number 496 - one of the known perfect numbers. In order to demonstrate that 496 is a perfect number, we must show that 496 = (the sum of all its divisors including 1 and excluding 496) We might just start by dividing and working out the divisors the long way. Or, we might begin by noting that, in the notation that includes a Mersenne prime, 496 = 24 (25 - 1) = 24 x 31. From this expression, we may easily obtain the divisors of 496, namely: 1, 2, 22, 23, 24, 31, (2×31), (22×31), (23×31), and (24×31). The sum of all the divisors excluding 496 is then 1 + 2 + 22 + 23 + 24 + 31× (1 + 2 + 22+ 23) In order to break this mess down a bit, let us examine the partial sum u = 1 + 2 + 22 + 23 + 24 If we multiply by 2 (i.e., 2u = 2 + 22 + 23 + 24 + 25) then subtract, we find u = 2u - u = 25 - 1 = 31 Thus, the sum of the divisors becomes = 31
+ 31(24 - 1) as we were to show.
nP = 2c(2c+1 - 1) be a perfect number with (2c+1 -1) being the embedded Mersenne prime. Then the divisors of nP are 1, 2, 22, ... , 2c, 2c+1 - 1, 2(2c+1 - 1), ..., 2c(2c+1 - 1) The sum of all the divisors excluding np is then 1 + 2 + 22 + ... + 2c + (2c+1 - 1) (1 + 2 + ... + 2c-1) Again, we examine the partial sum: u = 1 + 2 + 22 + ... + 2c. We multiply by 2 (i.e., 2u = 2 + 22 + ... + 2c + 2c+1) then subtract: u = 2u - u = 2c+1 - 1 The sum of the divisors becomes: = (2c+1
- 1) + (2c+1 - 1)(2c - 1) as we were to show.
|
||||
Please send suggestions/corrections to: Web Related: David.Mazza@grc.nasa.gov Technology Related: Joseph.C.Kolecki@grc.nasa.gov Responsible NASA Official: Theresa.M.Scott (Acting) |