I see this kind of problem has already appeared not long ago in the forums.
Well, the key is always to remove the fractions and factorize.
1) remove the fractions:
1/a+1/ab=1/2005 to remove all denominators we multiply both sides by 2005ab so we get 2005b+2005=ab
we put first the bs together 2005=ab-2005b and then we factorize b :
Now there has to be a shortcut. We can not ask you to find barehanded tens of integers. So we see that b multiplied by (a-2005) gives 2005 so b is a factor of 2005 and (a-2005) also is a factor of 2005.
So we search for the prime factorization of 2005 which is 2005=5*401 (I think that 401 is prime)
So we do not have many choices : either b=5 and so (a-2005)=401 or b=401 and (a-2005)=5 so we find "a" and we get two choices :
b=5 and a=2406
b=401 and a=2010
Always remove the fractions, factorize and then work with the factors .
Sorry I missed two :
b=1 and (a-2005) = 2005 so b=1 and a=4010
b=2005 and (a-2005) = 1 so b=2005 and a=2006