Question 1165362
        f(mn) = f(m)∙f(n) - f(m+n) + 1001
<pre>
Let m=1, then

        f(1n) = f(1)∙f(n) - f(1+n) + 1001
         f(n) = 2∙f(n) - f(n+1) + 1001
       f(n+1) = f(n) + 1001
f(n+1) - f(n) = 1001 

This tells us that after the first term the common difference between
a term and its preceding term equals 1001.

So it is an arithmetic sequence with first term f(1) = 2 and common
difference 1001.

     The formula is a<sub>n</sub> = a<sub>1</sub> + (n-1)∙d

                  f(n) = f(1) + (n-1)∙1001
                  f(n) = 2 + (n-1)∙1001
                  f(n) = 2 + 1001n - 1001
                  f(n) = 1001n - 999

So when n=12
               f(9999) = 1001(9999) - 999
               f(9999) = 10008999 - 999
               f(9999) = 10008000

Edwin</pre>