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) About Me  (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) About Me  (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

3%5E2008%2B4%5E2009=3%5E2008%2B4%2A4%2A2008=%283%5E502%29%5E4%2B4%284%5E502%29

We let x=3%5E502 and y+=4%5E502

Then


  

Substituting back for x and y

%28x%5E2-2xy%2B2y%5E2%29%28x%5E2%2B2xy%2B2y%5E2%29


  


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

3%5E1004-2%2A3%5E502%2A4%5E502%2B2%2A4%5E1004

is greater than 2009%5E182.

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