Angstrom Numbers

Date: 02/20/98 at 05:27:34
From: Anirban Basak
Subject: Angstrom Numbers

Hi Dr. Math! 

This question is on what I've heard about something called Angstrom 
numbers.

These are the numbers (0 and 1 excluded) in which the sum of the cube 
of the digits is equal to the number itself.
e.g. 153 = 1*1*1 + 5*5*5 + 3*3*3 = 153

Three more such numbers are 370, 371 and 407.

Is it true that between the numbers 2 and 10000, there are only 4 
Angstrom numbers? How do I know for certain? What is the next Angstrom 
number following 407?

Thanks.

Date: 02/20/98 at 13:47:54
From: Doctor Rob
Subject: Re: Angstrom Numbers

My computer verifies this that there are only the four you give up to 
10000.

Angstrom numbers cannot be very large. If a number has k digits, the 
sum of the cubes of its digits is at most 

     k*9^3 = 729*k.  

If a number has k digits, it is between 10^(k-1) and 10^k-1, 
inclusive.  Thus you would have to have 10^(k-1) <= 729*k < 10^k - 1.  
The second inequality is easy to satisfy, but the first one is not.  
It is false for k >= 5.  For that reason, the an Angstrom number can 
have only at most four digits.

To verify that there are no others with four or fewer digits, you can
proceed as follows. Pick four integers to be digits. You might as well 
make them in increasing order. Cube them, and add the cubes.  If you 
get a number with the original four digits, you have an Angstrom 
number. If not, there is no Angstrom number with those four digits in 
any order. You can run through all 210 possibilities with four 
different digits, plus all 360 with one pair, plus all 45 with two 
pairs, plus all 90 with three of a kind, plus all 10 with four of a 
kind. This is a total of 715 cases to consider (a lot better than 
10000 cases!). By allowing 0 to be a digit (even a leading digit), 
you are automatically taking care of the cases where the number has 
fewer than 4 digits.

I wish I could think of a better approach, but, alas, that's the best 
I can come up with.

-Doctor Rob,  The Math Forum
 http://mathforum.org/dr.math/