Question 836115
There are many ways to get to the answer.
I listed the one that seemed easier for my own mind to understand and calculate.
With some luck, it might sound logical and understandable for you too.
 
There are 5 odd digits: 1, 3, 5, 7, and 9.
 
How many 5-digit numbers can we make using those digits without repetition?
We have 5 ways to choose the first digit.
For each of those choices, there will be 4 ways to chose the second digit.
For each set of first and second digit choices, there will be 3 ways to chose the third digit.
For each set of first, second and third digit choices, there will be 2 ways to chose the fourth digit.
all in all, there are {{{5!=5*4*3*2=120}}} possible 5-digit numbers made from the 5 odd digits without repeating digits.
 
All 5 of those 5 odd digits are equally likely as the last (units) digit.
The last digit will be a 1 in {{{120/5=24}}} of those {{{120}}} numbers.
The last digit will be a 3 in another 24 numbers, a 5 in another 24, a 7 in another 24, and a 9 in another 24 numbers.
All the last digits added up will add up to
{{{24*(1+3+5+7+9)=24*25=600}}} .
 
The sum of the tens digits will also be {{{600}}} ,
and so will be the sums of the hundreds, thousands, and ten-thousands digits.
 
The value of the sum of all the numbers will be
{{{600*10000+600*1000+600*100+600*10+600="6 , 666 , 600"}}}