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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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up vote 7 down vote accepted

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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but let me ask,is my answer right ??,as i am not a person with formal training, – Iyengar Sep 7 '11 at 15:32
Why do you need this line? $24(1+3+5+7+9) = 24*25 = 600$. Why sum the numbers? I don't know the answer yet(trying not to look in the back of the book for the answers until I am sure I have this right).:) – user10695 Sep 7 '11 at 15:34
My question is, why add the individual allowable digits, rather than just adding the total occurrence for each? I was thinking more something like this $24(5) = 24*5 = 120$ – user10695 Sep 7 '11 at 15:35
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. – Iyengar Sep 7 '11 at 15:37

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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where did you get 55555 from? The question specifies that there are no repeats allowed. – user10695 Sep 7 '11 at 16:15
If you have for example $17935$ then you also have $111110-17935=93175$, so the average term is half $111110$, namely $55555$. – Henry Sep 7 '11 at 16:23
I am sorry, I still don't get the 111110 and the 55555. Why those? – user10695 Sep 7 '11 at 19:30

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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I need help seeing this [If $d$ is one of the allowable digits $1,3,5,7,9$, let $d' = 10 - d$; note that $d'$ is also allowable. ] please. Why is $10-d$ allowable? – user10695 Sep 7 '11 at 22:04
Also, how is $abcde + a'b'c'd'e' = 111110$. I don't get that please. Is this binary? – user10695 Sep 7 '11 at 22:06
@user10695: 1st question: Just do the arithmetic: if $d=1$, $10-d=9$; if $d=3$, $10-d=7$; and so on. Subtracting $1,3,5,7$, or $9$ from $10$ leaves $9,7,5,3$, or $1$, respectively. 2nd question: Not binary, just ordinary addition. $e+e'=10$, so you write down $0$ and carry $1$. Then $d+d'=10$, and the carry makes $11$, so you write down $1$ and carry $1$. The same thing happens in the remaining three columns, so you end up with a total of $111110$. – Brian M. Scott Sep 7 '11 at 22:15
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. – user10695 Sep 7 '11 at 22:33

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. – SchrodingersCat Oct 28 '15 at 16:30

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