decorative picture of space events
Aeronautics Home PageVirtual Visits (Video Conferencing, Virtual Tours, Webcasts, plus more.)Math and Science ResourcesTeachers ResourcesInternet Access Research ArchiveLink to Learning Technologies Project (LTP) Homepage

       Math & Science Home | Proficiency Tests | Mathematical Thinking in Physics | Aeronauts 2000

CONTENTS
 

Introduction

Fermi's Piano Tuner Problem

How Old is Old?

If the Terrestrial Poles were to Melt...

Sunlight Exerts Pressure

Falling Eastward

What if an Asteroid Hit the Earth

Using a Jeep to Estimate the Energy in Gasoline

How do Police Radars really work?

How "Fast" is the Speed of Light?

How Long is a Light Year?

How Big is a Trillion?

"Seeing" the Earth, Moon, and Sun to Scale

Of Stars and Drops of Water

If I Were to Build a Model of the Cosmos...

A Number Trick

Designing a High Altitude Balloon

Pressure in the Vicinity of a Lunar Astronaut Space Suit due to Outgassing of Coolant Water

Calendar Calculations

Telling Time by the Stars - Sidereal Time

Fields, an Heuristic Approach

The Irrationality of

The Irrationality of

The Number (i)i

Estimating the Temperature of a Flat Plate in Low Earth Orbit

Proving that (p)1/n is Irrational when p is a Prime and n>1

The Transcendentality of

Ideal Gases under Constant Volume, Constant Pressure, Constant Temperature and Adiabatic Conditions

Maxwell's Equations: The Vector and Scalar Potentials

A Possible Scalar Term Describing Energy Density in the Gravitational Field

A Proposed Relativistic, Thermodynamic Four-Vector

Motivational Argument for the Expression-eix=cosx+isinx

Another Motivational Argument for the Expression-eix=cosx+isinx
Calculating the Energy from Sunlight over a 12 hour period
Calculating the Energy from Sunlight over actual full day
Perfect Numbers-A Case Study
Gravitation Inside a Uniform Hollow Sphere
Further note on Gravitation Inside a Uniform Hollow Sphere
Pythagorean Triples
Black Holes and Point Set Topology
Additional Notes on Black Holes and Point Set Topology
Field Equations and Equations of Motion (General Relativity)
The observer in modern physics
A Note on the Centrifugal and Coriolis Accelerations as Pseudo Accelerations - PDF File
On Expansion of the Universe - PDF File
 
Perfect Numbers - A Case Study

Perfect 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)
= 24×31
= 496

as we were to show.
To show that any perfect number may be broken down in this way, we let

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)
= 2c(2c+1 - 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)