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

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To determine the sum of all four-digit numbers formed by the digits 1, 3, 5, and 7 without repetition, a systematic approach is suggested rather than listing all combinations. The key is to analyze how often each digit appears in each positional place (thousands, hundreds, tens, and units) across all permutations. Understanding that each digit contributes to the overall sum based on its positional value can lead to a general formula for larger sets of digits. The discussion emphasizes the importance of breaking down the problem into manageable parts, possibly starting with fewer digits to grasp the concept better. This method allows for a more efficient calculation without exhaustive enumeration.
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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?
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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