Solve Large Numerical Values Without a Calculator

[PageWizard]

Hello everyone :),

I'm sure someone will have no problem helping me with this. How do I figure out large powered numerical values without using a calculator. Kept getting questions involving 2^2007 and others. How would I be able to figure out these values without using a calculator or a lot of time on my hands? I haven't studied yet how to figure this out. Hope that makes sense, and thank you for your time to helping me with this :).

 

[Redbelly98]

I'm not certain I understand just what you want. It seems pretty pointless to be able to write out all the digits in the number 2^2007. So how about an approximate result where you find the nearest power of 10?

Basically, you need to know that log(2) is approximately 0.3. Using that, we can write:

<br /> 2^{2007} = 10^{\log(2^{2007})}<br /> = 10^{2007 \cdot log(2)}<br /> \approx 10^{2007 \ \cdot \ 0.3}<br /> \approx 10^{602}<br />

Or ... you can use the fact (familiar to those knowledgeable about computers) that 2^10 is approximately 1000 or 10^3 (It's really 1024, but we're approximating here). So:

<br /> 2^{2007} = 2^{10 \ \cdot \ 200.7} = (2^{10})^{200.7} \approx (10^3)^{200.7} = 10^{3 \ \cdot \ 200.7} \approx 10^{602}<br />

 

[CRGreathouse]

Or...

the questions may have been about modular arithmetic, like "what is the last digit of 2^2007". In that case, you look at

2^1 = 2 (mod 10)
2^2 = 4 (mod 10)
2^3 = 8 (mod 10)
2^4 = 6 (mod 10)
2^5 = 2 (mod 10)
. . .

and notice that the pattern repeats.

 

[PageWizard]

I think they would be looking for modular arithmetic since this is a number theory style proof they request for these type of questions. But, this helps me out thank you, gives me somewhere to start :).