Difference of number and reverse number is 84942. What would be the sum of all such possible digits

Question

A five digit number has all non-zero different digits & decreasing arranged. Average of all the digits of the numbers is 5. The difference of number and the number formed by reversing the order of digits is 84942.

What would be the sum of all such possible digits -
1) 196052
2) 196002
3) 195984
4) 195842
5) 195802
6) 195798
7) 195602
8) 195588
9) 195512
10)195490

Answer 1

Let the no. be abcde where a>b>c>d>e
Given = a+b+c+d+e = 25
and a b c d e
-e d c b a
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
8 4 9 4 2
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
Solve it using cryptographic techniques
1st digit a-e=8,, so take a=9 and e=1
using this d-1+10-b=4 { second last digit }
d+5=b
now take d=2 , b=7 or d=3 , b=8
putting all the values and c a random we get
a= 9 a=9
b= 7 b=8
c= c=
d= 2 d=3
e=1 e=1
Now using given condition a+b+c+d+e=25;;
c=6 c=4
Hence nos are 97621 and 98431 and no other case satisfies the conditions in the question ,
Hence sum = 97621
+98431
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196052

Answer 2

(A1*10^4+A2*10^3+A3*10^2+A4*10+A5) - (A5*10^4+A4*10^3+A3*10^2+A2*10+A1) = 84942
1111*(A1 - A5) + 110*(A2 - A4) = 9438
Now: becuase A1 - A5 is 8 hence A1 and A5 can only be 9 and 1
therefore: 110(A2 - A4)= 550
and hence: A2 - A4 = 5
A3 can be any number between A2 and A4
The result would therefor be: 98731, 98631, 98531, 98431, 97621, 97521, 97421, 97321