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

- ParthKohli

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

- chestercat

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- ParthKohli

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.

- ParthKohli

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

- ParthKohli

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

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## More answers

- ParthKohli

Something with a pattern?

- shubhamsrg

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.

- terenzreignz

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

- shubhamsrg

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

- ParthKohli

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

- terenzreignz

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

- terenzreignz

then multiply 39 to it again.

- terenzreignz

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

- ParthKohli

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

- shubhamsrg

Ohh, I see.

- terenzreignz

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.

- terenzreignz

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

- ParthKohli

Nice... can you give me another example?

- terenzreignz

How about
6^5000 = n (mod 215)

- terenzreignz

(sneakily thinking of simple problems) :>

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- terenzreignz

I'm not following o.O
You may be using a method I'm not familiar with, though... if you are, do continue :)

- ParthKohli

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

- ParthKohli

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

- ParthKohli

What should I do here?

- terenzreignz

Hang on...

- terenzreignz

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

- ParthKohli

Oh...

- terenzreignz

Wait, typo
-.-

- ParthKohli

Yes, isn't that 1 (mod 215)

- terenzreignz

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

- terenzreignz

And work from there...

- terenzreignz

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

- ParthKohli

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

- terenzreignz

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

- ParthKohli

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

- ParthKohli

Argh, you type fast. =3

- terenzreignz

Awesomeness

- ParthKohli

Another n00bish question coming up. Thanks man.

- terenzreignz

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

- ParthKohli

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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