A community for students.
Here's the question you clicked on:
 0 viewing
ParthKohli
 2 years ago
A common technique for finding modulo \(10\) of any huge number in the form \(a^b\)?
ParthKohli
 2 years ago
A common technique for finding modulo \(10\) of any huge number in the form \(a^b\)?

This Question is Closed

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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.