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I have yet another analytical question that got me

A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?

Please kindly explain your final answer so those of us who are still trying to learn can grasp the logic easily. Thanks in advance!

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The number of possible combinations is $5! = 120$. Each digit will be in each position an equal number of times, namely $5!/5 = 24$ times.

So the sum of the digits in the one's place is

$24(1+3+5+7+9) = 24*25 = 600$

This will be the sum of the digits in each place (one's, ten's, etc.). So the total sum is:

$600*11,111 = 6,666,600$, regards, iyengar

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    For each decimal place, 24 of these are of the same digit. Thus for a decimal place the total value is 24 (1 + 3 + 5 + 7 + 9) = 600 To get the total value of all these numbers multiply the first result by 11111.2011-09-07
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As a slight alternative to iyengar's answer:

There are $120$ such numbers and their average is $55555$ (note that for any of the form, there is also $111110$ minus that number) so the answer is $120\times 55555=6666600$.

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    I am sorry, I still don't get the$111110$and the 55555. Why those?2011-09-07
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This is partly a slight variant of Henry’s answer and partly a further explanation in answer to one of your questions.

If $d$ is one of the allowable digits $1,3,5,7,9$, let d' = 10 - d; note that d' is also allowable. Now let $abcde$ be any allowable number; then a'b'c'd'e' is also allowable, and it must be a different number from $abcde$. (It would be the same number only if all of the digits were $5$, but that’s not allowed.) In this way you can pair up all of the allowable numbers. Since there are $5! = 120$ permutations of the $5$ allowable digits, there are $120$ allowable numbers, and hence there are $60$ of these pairs.

It shouldn’t be too hard to see that abcde + a'b'c'd'e' = 111110 no matter which allowable number $abcde$ you start with. That is, the numbers in each pair sum to $111110$. Since there are $60$ pairs, the grand total is $60 \cdot 111110 = 6,666,600$.

Henry did essentially the same thing, except that instead of looking at the sum of $abcde$ and a'b'c'd'e', he looked at their average, $111110/2 = 55555$. Since all pairs have the same average, the entire collection of allowable numbers must also have that average. And if the $120$ allowable numbers have an average of $55555$, their total must be $120 \cdot 55555 = 6,666,600$.

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    Thanks! What I thought $abcde + a'b'c'd'e' = 111110$ meant was $(a*b*c*d*e) + (a'*b'*c'*d'*e')$ . Now I see what you mean. Thanks for that.2011-09-07
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why all this ?

it is so simple There are 5 digits and as we all know it will 5! numbers (1*2*3*4*5)=120 on a digit place every number will repeat 24 times that is 120/5 on ones place the sum will be 24(1+3+5+7+9)=25*24=600 If u need further explanation why 24(1+3+5+7+9+) see this: at every digit place 1 will repeat 24 times 24*1=24 3 will repeat 24 times 24*3=72 5 will repeat 24 times 24*5=120 7 will repeat 24 times 24*7=168 9 will repeat 24 times 24*9=216 total give 600

every digit place will sum to 600 unit place 600 tenth place 600 hundreds place 600 Thousands place 600 ten thousands place 600 -------------------------------- 6666600

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    Use MathJax please.2015-10-28