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OK, I'm gonna change the topic to something else:\[39^{39} \equiv n\pmod{100}\]

\(n\) has to be a two-digit number.
(I actually have to find the last two digits)

Something with a pattern?

Better yet,
39^5 = -1 mod 100
Use that:
39^39 = (39^5)^7 * 39^4 = (-1)^7 * 39^4 mod 100

Seems legit.
But on paper, who's gonna go till 39^5 ! :O

Very nice, can you tell me *how* you came up with that solution? What do you initially think?

square 39, and then only take the last two digits...

then multiply 39 to it again.

and only take the last two digits, again...

Ahem, that was beautiful. But is there a common way to figure out these solutions?

Ohh, I see.

That "last two digits" trick only works because you were conveniently working under mod 100... XD

Nice... can you give me another example?

How about
6^5000 = n (mod 215)

(sneakily thinking of simple problems) :>

Arhmegurd.\[\left(6^{5}\right)^{1000} \equiv (6^2)^{1000} \equiv n \pmod{215}\]

\[\left(6^5\right)^{400} \equiv (6^2)^{400} \equiv n\pmod{215}\]

\[\left(6^5\right)^{120} \equiv 6^{240} \equiv n \pmod{215}\]

\[\left(6^{5}\right)^{48} \equiv 6^{96} \equiv n \pmod{215}\]

Hmm... I am just using \(6^2 \equiv 6^5 \pmod{215}\)

What should I do here?

Hang on...

I was rather hoping you'd go for
\[6^3 = 216 = -1(mod \ 215)\]

Oh...

Wait, typo
-.-

Yes, isn't that 1 (mod 215)

\[\large 6^3 = 216 = 1(\mod \ 215 \ \ \ \ )\]
Much better

And work from there...

\[\large 6^{5000} = 6^{3(1666)+2}\]

\[6^{5000}= 6^{4998} \times 36 \equiv \]

\[1 \times 36 (\mod 215 \ \ \ \ \ )\]

\[\equiv 36 \pmod{215}\]

Argh, you type fast. =3

Awesomeness

Another n00bish question coming up. Thanks man.

If you have time, look up Fermat's Little Theorem, might come in handy in a pinch.

Yesh, that one is cool:\[p^{n - 1} \equiv 1 \pmod{n} \]I'd try to see if that applies in the future.