in Combinatory
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Ans: 43333

in Combinatory
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We know five digit numbers are of the form $abcde$ and in particular the numbers are $a_1b_1c_1d_1e_1, a_2b_2c_2d_2e_2,...$

Sum of all such numbers will be $= \sum_{i\ge1} \left( a_i*10^4 + b_i*10^3+ c_i*10^2+ d_i*10^1 +e_i*10^0\right)$

$= ((2+3+5+7) * 4! * 10^4) + ((2+3+5+7)*3*3!*10^3) + ((2+3+5+7)*3*3!*10^2) + ((2+3+5+7)*3*3!*10^1) + ((2+3+5+7)*3*3!*10^0)$

$= (17*3!)(40000+3000+300+30+3)$

$= (17*3!)(43333)$

(.) How did I come up with $(2+3+5+7)*4!*10^4$ as one term in the summation?

$\rightarrow$ How many five digit numbers of the form $2bcde$ are there? $\rightarrow 4!$

Similarly, we have $4!$ five digit numbers starting from each of these digits $3,5,7$

Of all these numbers, if we only count the contribution of digit at $(10^4)^{th}$ place to our summation we'll get $(2+3+5+7)*4!*10^4$.

Similarly, other terms were obtained.

$S =(17*3!)(43333)$

$\therefore \frac{S}{102} = 43333$.

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to find the sum of all the possible 5 digit numbers 

we can use (5!/5)*(sum of digits)*(11111)

but as here 0 is also present so we have to subtract all the four digit numbers possible

i.e  (4!/4)*(sum of digits)*(1111)

then we will get the S 

after dividing it by 102 we will get 43333.