SOLUTION: show that 3^2008+4^2009 can be written as product of two positive integers each of which is larger than 2009^182.

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Question 252126: show that 3^2008+4^2009 can be written as product of two positive integers each of which is larger than 2009^182.
Found 2 solutions by drk, Edwin McCravy:
Answer by drk(1908)   (Show Source): You can put this solution on YOUR website!
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.
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Note: The tutor "drk" did not show that
3^2008[1 +3*(4/3)^2009] is the product of two integers. In fact
[1 +3*(4/3)^2009] is not an integer!!!

show that 3^2008+4^2009 can be written as product of two positive integers each of which is larger than 2009^182.

We use the fact that



We let  and 

Then


  

Substituting back for  and 




  


So this is the product of two very large integers. So we have

shown the first part.

Now we only need to show that the smaller of these, which is



is greater than .

I have figured out how to do this yet, but I'm working on it.

If and when I get it then I'll post it here.

Edwin
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