Sum of 4 Digit Numbers Using 1,3,5,7

  • Thread starter Thread starter hartigan83
  • Start date Start date
  • Tags Tags
    Formula
hartigan83
Messages
19
Reaction score
0
Is there a formula?

Homework Statement


Determine the sum of all the four digit numbers formed by using the digits 1,3,5, and 7. NO DIGIT REPEATED in any number.


Homework Equations


thats my question...


The Attempt at a Solution



I can solve this problem by simply writing out all of the terms and adding them up...
ie
1357+1375+...3157+3175+...+ 5137+5173+...+7135+7153+...7531 =:zzz:

But is there a formla which I can use for these types of questions? For example if there were 9 different numbers I would not want to write out all the combinations and then add up all the numbers. It wasnt to difficult with just 4 but, I knowthere must be some formula or theorem out there... Any ideas?
 
Physics news on Phys.org
I don't think there is such a formula, but here would be my idea (which in a sense could lend a general formula, but I think it would be ugly to write out explicitly).

How many times does 1 show up as the 1's digit (how about 10's digit, 100's digit, 1000's digit?)? How many times does any of the four digits (1,3,5,7) show up in any of the four possible positions (the 1's digit, 10's, 100's, 1000's)?

Once you know that, think about how the following might help you:

Any four digit number can be written as a sum of the following type: a10^3 + b10^2 + c10 + d where a,b,c,d are integers.

For example 1579 = 1000 + 500 + 70 + 9. (here a = 1, b + 5, c + 7 and d = 9).

Does that give any idea of how to do this problem in a more general way?
 
mattmns is thinking the same way I am as well on the subject. Look at all the possible combinations of those 4 numbers in a 4 digit number and examine each digit by itself over all the 4 digit numbers you made.

If 4 digits was too much, you might want to consider starting smaller with say the numbers 1, 3 and 7 combining to make a 3 digit number or 1 and 3 to make a 2 digit number.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top