Great, thanks a lot... but at this moment.. make some video on ICPC roadmap. Which topic we should learn. Also if possible, provide some resources or you can share your journey.. please
I think I got it. first, because x is a prime number, x^anything will only divide numbers that are powers of x, like x, x^2 x^3,... etc. also: We can factor x^(s−an) out of the numerator: x^(s−an) (1+x^(an−a1)+x^(an−a2)+⋯+x^(an-an−1)). so x^(s-an) is the largest number in the numerator that divides it without a remainder.
@YogeshRathee-v5d The x^(s-an) is considered gcd because it the least positive number in the series and we can take common as x^(s-an) so that the numerator and denominator is divisible...
Thank you so much Shayan! You teach excellently, and are also very humble. Looking forward to more sessions! :)
Thank you for your support. It means a lot. ❤️
Thanks a lot, Shayan. This series is very helpful & salute to your teaching style: 10/10
The proof of fermat little theorem is very interesting, never thought that way. Thanks for the explanation.
Great, thanks a lot... but at this moment.. make some video on ICPC roadmap. Which topic we should learn. Also if possible, provide some resources or you can share your journey.. please
Thanks a lot😊
thanks a ton
Hey! Great video, but I think you put the wrong problems in the description.
7/10
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can anyone explain me how x to the power of s - an is the gcd i tried thinking for an hour but couldn't get anywhere with it
Same here, I cannot see why this assumption was valid.
I think I got it.
first, because x is a prime number, x^anything will only divide numbers that are powers of x, like x, x^2
x^3,... etc.
also:
We can factor x^(s−an) out of the numerator:
x^(s−an) (1+x^(an−a1)+x^(an−a2)+⋯+x^(an-an−1)).
so x^(s-an) is the largest number in the numerator that divides it without a remainder.
@YogeshRathee-v5d The x^(s-an) is considered gcd because it the least positive number in the series and we can take common as x^(s-an) so that the numerator and denominator is divisible...