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A common technique for finding modulo \(10\) of any huge number in the form \(a^b\)?

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It seems that you have to go through a lot of manipulations... a lot for doing that. Although I was thinking of a technique they use to figure all that out.
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)

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Other answers:

Something with a pattern?
39 = -61 mod100 =>39^38 = 61^38 mod100 =>39^39 = 39*61^38 mod100 SInce 61^n always ends in 1, that multiplied by 39, will give last digit as 9, hmm, leme see for the second last.
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.
The way I think of things, and it's a common trend that the way I do it is highly inefficient (LOL) But when faced with a problem b^p = n (mod k) I try to see if there's an exponent to which b can be raised so that it is 1 (or -1) mod k first.
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.\[\left(6^{5}\right)^{1000} \equiv (6^2)^{1000} \equiv n \pmod{215}\]
\[\left(6^5\right)^{400} \equiv (6^2)^{400} \equiv n\pmod{215}\]
\[\left(6^5\right)^{120} \equiv 6^{240} \equiv n \pmod{215}\]
I'm not following o.O You may be using a method I'm not familiar with, though... if you are, do continue :)
\[\left(6^{5}\right)^{48} \equiv 6^{96} \equiv n \pmod{215}\]
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 \[6^3 = 216 = -1(mod \ 215)\]
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
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.

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