Finding the Last Digit of 2009^2009

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1. What is the last digit of 2009^2009



I think you go about this by factoring 2009 as 7*7*41. I'm pretty much stuck.
 
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What are the last digits of 2009^1, 2009^2, 2009^3 etc? Keep going until you see the pattern.
 
To simplify dick's post: it suffices to look at 9^1, 9^2, 9^3,..., do you see why?
 
micromass said:
To simplify dick's post: it suffices to look at 9^1, 9^2, 9^3,..., do you see why?

And it's even easier than that if you know mod arithmetic and notice 2009 is equal to (-1) mod 10.
 
yes thank you, i see the pattern.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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