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

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0OK, I'm gonna change the topic to something else:\[39^{39} \equiv n\pmod{100}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\(n\) has to be a twodigit number. (I actually have to find the last two digits)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Something with a pattern?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.039 = 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
 one year ago
Best ResponseYou've already chosen the best response.2Better yet, 39^5 = 1 mod 100 Use that: 39^39 = (39^5)^7 * 39^4 = (1)^7 * 39^4 mod 100

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0Seems legit. But on paper, who's gonna go till 39^5 ! :O

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Very nice, can you tell me *how* you came up with that solution? What do you initially think?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2square 39, and then only take the last two digits...

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2then multiply 39 to it again.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2and only take the last two digits, again...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Ahem, that was beautiful. But is there a common way to figure out these solutions?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.2That "last two digits" trick only works because you were conveniently working under mod 100... XD

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Nice... can you give me another example?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2How about 6^5000 = n (mod 215)

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2(sneakily thinking of simple problems) :>

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Arhmegurd.\[\left(6^{5}\right)^{1000} \equiv (6^2)^{1000} \equiv n \pmod{215}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(6^5\right)^{400} \equiv (6^2)^{400} \equiv n\pmod{215}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(6^5\right)^{120} \equiv 6^{240} \equiv n \pmod{215}\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2I'm not following o.O You may be using a method I'm not familiar with, though... if you are, do continue :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(6^{5}\right)^{48} \equiv 6^{96} \equiv n \pmod{215}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Hmm... I am just using \(6^2 \equiv 6^5 \pmod{215}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0What should I do here?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2I was rather hoping you'd go for \[6^3 = 216 = 1(mod \ 215)\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Yes, isn't that 1 (mod 215)

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2\[\large 6^3 = 216 = 1(\mod \ 215 \ \ \ \ )\] Much better

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2And work from there...

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2\[\large 6^{5000} = 6^{3(1666)+2}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[6^{5000}= 6^{4998} \times 36 \equiv \]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2\[1 \times 36 (\mod 215 \ \ \ \ \ )\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\equiv 36 \pmod{215}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Argh, you type fast. =3

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Another n00bish question coming up. Thanks man.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2If you have time, look up Fermat's Little Theorem, might come in handy in a pinch.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Yesh, 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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