Is 120 a Perfect Number?
Date: 10/15/97 at 12:00:39 From: Chris Sweigard Subject: Perfect numbers Is 120 a perfect number? If not, why not? It seems to fit all of the criteria. Chris
Date: 10/15/97 at 13:53:42 From: Doctor Wilkinson Subject: Re: Perfect numbers No, it isn't. The divisors of 120 other than 120 are 1 2 4 8 3 6 12 24 5 10 20 40 15 30 60 If you add these up you get 240, not 120. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 10/21/97 at 14:16:39 From: Doctor Mark Subject: Re: Perfect numbers Hi Chris, As I am sure you know, a perfect number is a number which is equal to the sum of its proper divisors (all the numbers, including 1, that divide the number, except for the number itself). So what are the proper divisors of 120? This can be done systematically, from just looking at all the numbers from 1 to 1/2(120) (= 60) that divide 120 (do you see why we don't have to look at any numbers greater than 60?). You could do this using a calculator in a few minutes, but it is a little tedious, so you might try doing it systematically in another way, using the prime factorization of 120. Here's how that goes. First, find the prime factorization of 120: 2 x 2 x 2 x 3 x 5. Now suppose that a number d is a divisor of 120. How can we figure out all the divisors of 120 using this prime factorization? If you think about it a little bit, you can see that the divisors of d have to have a prime factorization of the form: d = (no more than three 2's) x (no more than one 3) x (no more than one 5) with no other primes in its factorization [I explain this at the end of this note]. But then it is pretty simple to see what all the divisors of 120 can be. They have to be of the form: d = (1, 2, 4, or 8) x (1 or 3) x (1 or 5). Consequently, we can systematically list all the possibilities: the product (1 or 3) x (1 or 5) can give either 1 x 1 = 1, or 1 x 5 = 5, or 3 x 1 = 3, or 3 x 5 = 15, i.e., we can get 1, 3, 5, or 15. If we multiply each of these numbers by 1, 2, 4, or 8, we will get all the divisors of 120: 1 x (1, 3, 5, or 15) = 1, 3, 5, 15 2 x (1, 3, 5, or 15) = 2, 6, 10, 30 4 x (1, 3, 5, or 15) = 4, 12, 20, 60 8 x (1, 3, 5, or 15) = 8, 24, 40, 120 The divisors of 120 are then only the following: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. If we add all of these *except* 120 (remember that the number itself is not a *proper* divisor!) together, we get 240, which is twice the number we started out with, 120. So 120 is not perfect. However, you have discovered that 120 is what is called a "3-perfect" number: a number whose sum of proper divisors is k times the number is called "k+1-perfect" (a regular old perfect number is a "2-perfect" number). [The reason for this "k+1" stuff is that if you allow the number itself as a divisor, then a perfect number can be defined as a number whose sum-of-divisors is *twice* the number itself. That's why we define a regular old perfect number as "2-perfect."] There are six other known numbers which, like 120, are 3-perfect. One of these is the number 459,818,240 (the sum of its divisors is twice the number itself). Using computers, there have even been found examples of numbers which are 9-perfect, i.e., the sum of the proper divisors of the number is 8 times the number itself. I'd tell you what it is, but I couldn't write it down without my hands falling off. Suffice it to say that it has 38 different primes in its prime factorization, with 114 factors of 2 and 35 factors of 3, among other things. The largest prime in its prime factorization is 2,646,507,710,984,041! (Don't try to show that, though, unless you have an incredible amount of free time on your hands or a fast computer.) According to my sources (c. 1993), no one has yet discovered a 10-perfect number, though people generally believe that there are k-perfect numbers no matter what you choose for k. But maybe someone has discovered one and just hasn't bothered to tell me yet... Perfect numbers that are even are related to numbers called Mersenne primes, and one can show that for every Mersenne prime, there is a single even perfect number, and every even perfect number comes from such a Mersenne prime. The latest Mersenne prime was found in September, and has about 895,000 digits! There's a lot of neat stuff related to Mersenne primes, and you can find out more than you probably ever wanted to know on the WWW. For more information, see: http://www.mersenne.org/ No one has ever found a perfect number that is odd. We do know, however, that if such an odd perfect number exists, it must have more than 100 digits. So now, how about my earlier statement about the possible divisors of 120? It's easiest to see that if you think of fractions. Remember that the division of A by B is represented by the fraction A/B. So if d divides 120, then we must have that 120 divided by d is an integer, i.e., the fraction 120/d must be an integer. If you think about the prime factorization of 120, and the prime factorization of d, you see that the only way that we can get 120/d to be an integer is if all the primes in the prime factorization of the bottom (d) of this fraction have an evil twin on the top (120), with which they can cancel. This means that only the primes 2, 3, and 5 can appear in the prime factorization of d, and there can't be any more of them in this prime factorization than there are in the prime factorization of 120. If either of these statements were false, then we would end up with a factor on the bottom that did not get cancelled off, and there is no way we could get an integer for 120/d. Be sure to write back if you have any questions about my explanation! -Doctor Mark, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum