Question 75181
Suppose 34a47b5893 is divisible by 99
Then obviously it is divisible by 9
If a number is divisible by 9 the sum of all the digits should divisible by 9 (this can be easily proved)

For example let the number abcd divisible by 9 then
{{{10^3a+10^2b+10c+d = 9k}}} where k is positive integer 
Then {{{1000a+100b+10c+d = 9k}}}
{{{(999a+a)+(99b+b)+(9c+c)+d=9k}}}
{{{a+b+c+d = 9k-(999a+99b+9c)}}}

{{{a+b+c+d = 9(k-111a+11b+c)=9K}}}
then a+b+d+c is divisible by 9
This can be proved for any number (can be given a general proof. Try it)

Again to the problem 
Then the sum of 34a47b5893 is divisible by 9

Then 3+4+a+4+7+b+5+8+9+3 is divisible by 9
Then
43+a+b is divisible by nine 

Then a=0 and b=2
Or a=2 and b = 0