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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

terenzreignz
 2 years 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
 2 years 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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