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

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ParthKohliBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

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

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

shubhamsrgBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

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

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

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

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

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

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

terenzreignzBest ResponseYou've already chosen the best response.2
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.
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ParthKohliBest ResponseYou've already chosen the best response.0
Yesh, that one is cool:\[p^{n  1} \equiv 1 \pmod{n} \]I'd try to see if that applies in the future.
 one year ago
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