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Is there a formula?

  1. Mar 4, 2007 #1
    Is there a formula????

    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    thats my question......

    3. The attempt at a solution

    I can solve this problem by simply writing out all of the terms and adding them up....
    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?
  2. jcsd
  3. Mar 4, 2007 #2
    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: [itex]a10^3 + b10^2 + c10 + d[/itex] 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?
  4. Mar 4, 2007 #3
    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.
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