Question 252126
First, lets rewrite 4^2009 as (3*(4/3))^2009. 
Now, we have 3^2008 + [3(4/3)]^2009.
A GCF is 3^2008, giving us 3^2008[1 +3*(4/3)^2009].
Now the tricky part. To find how many digits a number of the form a^b has, we have the formula D = [bloga] + 1, where D = number of digits and [ ] is the floor function, or the smallest integers less than or equal to the given number.
For 3^2008, D = [3*log(2008)] + 1 ~ [958.06] + 1 = 958 + 1 = 959 digits.
For 3*(4/3)^2009, D = 3[(4/3)*log(2009)] + 1 ~ 753 + 1 = 754 digits.

Now, we turn to 2009^182 and apply the same formula. D = [182*log(2009)] + 1 ~ 601 + 1 = 602 digits.

Clearly, we have shown that 3^2008+4^2009 can be written as product of two positive integers each of which is larger than 2009^182.