| Author | Maths question |
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You dont have to solve this finding the value, that's dumb
There's some other way to do this and that's what I wanna know |
Which chapter or unit does this question belongs to?
I may be able to help if I get to know the portion from which it comes from. :) |
| These are the kind of questions asked in aptitude rounds of data analytics company recruitment or in any competitive exams like CAT/GMAT |
Which chapter or unit does this question belongs to?
I may be able to help if I get to know the portion from which it comes from. :)
No chapter of mine :D , I just found it and it stuck in my head |
I guess you should try it with binomial. I am trying. Will let you know if I get to answer.
You know, it's bit tough for someone from bio side. :P
But no problem in giving it a try ;) |
Wouldn't that be a problem about the least common multiple?
Learned that too long ago, I've forgotten how it works but that's what comes to mind. |
| Nope, that's different. I'm sure, there's some binomial play here :) |
Nope, that's different. I'm sure, there's some binomial play here :)
No
Wouldn't that be a problem about the least common multiple?
Not LCM but I think this can be done using periodicity which I'm also not sure how to apply |
Maybe PM randomr :)
He's good at math stuff |
| this is easy, it is a direct application of the euler theorem. if you dont know the theorem just go to any math website or book to understand it and then you can one shot this problem in your head. no pen or paper needed. if you dont get how to solve it once you have studied the theorem just message me and i will help you. |
Let x = (8^828) mod 167 = 2^(3*828) mod 167 = 2^(14*166 + 160) mod 167 = (2^166)^14 * 2^160 mod 167.
By Fermat's little theorem 2^166 = 1 mod 167, because 167 is prime (if 167 were not prime you could use Euler's Theorem as per Igles' suggestion)
Thus, x = 2^160 mod 167 = 2^(2^5*5) mod 167 = (2^5)^(2^5) mod 167 = 32^32 mod 167 = (32^2)^16 mod 167 = (32^2 mod 167)^16 mod 167 = 22^16 mod 167 = (22^2 mod 167)^8 mod 169 = 150^8 mod 167 = (150^2 mod 167)^4 mod 167 = 122^4 mod 167 = (122^2 mod 167)^2 mod 167 = 21^2 mod 167 = 107. |
| closed by Bunnie (2020-01-02 09:58:25) |
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