@@MathWisdom42 I got 6 because we first look at divors of 2004 which are 2^2(3)(167), now we se that we facor put 2 from eaach of the numbers, and those numbers would be odd, so there only one 2 in GCD. Now we 3 and 167, now we see that 167 does not divide 2002^2+2 giving us the idea that GCD is 2 or 6, so we have find if 3 divides all the things in the set. Now using some modular arthmatic and pattren finding we can find that 3 divides it( I am tryint to post a link to my work but youtube is hiding it), so 6 is GCD.
By algebra I found that n^17 - n = n(n^8 + 1)(n^4 + 1)(n^2 + 1)(n+1)(n-1) = M. Since n,n+1 and n-1 are all factors, M is divisible by 3 and 2. Using Fermat's little theorem, n^17 == n mod 17. Therefore, n^17 - n == n - n == 0 mod 17. So 17 divides M. Also, M = (n^8 + 1)(n^4 + 1)(n^5 - n), and n^5 - n == n - n == 0 mod 5. So 5 divides M. (We could again show that 3 and 2 are divisors because M's is a multiple of (n^2 - n) and of (n^3 - n), and n^2 - n == n - n == 0 mod 2, and n^3 - n == 0 mod 3.) Thus we have shown that 2, 3, 5 and 17 are always divisors of M. And 257 is not always a divisor. The GCD is 510.
Explain the problem more. We are looking for the greatest common divisor of n^17 - n and n? It would have to be n. Are we looking for the greatest common divisor of n^17 and n? It would also be n. Are we looking for the greatest number that is a gcd of n^17 - n and any real number? It would be n^17 - n. Or maybe we are just looking for the greatest divisor of n^17 - n. No "common". But the greatest divisor of n^17 - n would still be n. So we are really looking for the greatest number d that divides n^17 - n for any n.
Yeah, that was my first reaction, what is "gcd" doing in the first statement of the problem???? It would be better stated as "find the largest k in N such that k is a divisor of n^17-n for all n in N"
To be fair, Andrew Tate sounds good. He would be considered to be an excellent success if only he took back all the thing he said about money, people, society, etc.
1.) We know it must be a product of primes to the first power. It cannot contain a square, because p^2 wouldn't divide p^17-p. 2.) It suffices to check for prime factors less than 17, because otherwise we know that there exists a primitive root of p, therefore an n such that p is the smallest number satisfying p | (n^p-n) therefore p doesn't divide n^17-n.
Anyone managed to solve the problem at the end?
Is the answer 6?
Great ! Feel free to share your reasoning@@JoE-mAmA-yO-mAmA
@@MathWisdom42 I got 6 because we first look at divors of 2004 which are 2^2(3)(167), now we se that we facor put 2 from eaach of the numbers, and those numbers would be odd, so there only one 2 in GCD. Now we 3 and 167, now we see that 167 does not divide 2002^2+2 giving us the idea that GCD is 2 or 6, so we have find if 3 divides all the things in the set. Now using some modular arthmatic and pattren finding we can find that 3 divides it( I am tryint to post a link to my work but youtube is hiding it), so 6 is GCD.
Well done ! @@John-cl8iv2
thanks @@MathWisdom42
this wasnt supposed to go THIS hard
Huge w bro, we seriously need more of these.
This is so oddly motivating
By algebra I found that n^17 - n = n(n^8 + 1)(n^4 + 1)(n^2 + 1)(n+1)(n-1) = M.
Since n,n+1 and n-1 are all factors, M is divisible by 3 and 2.
Using Fermat's little theorem, n^17 == n mod 17. Therefore, n^17 - n == n - n == 0 mod 17. So 17 divides M.
Also, M = (n^8 + 1)(n^4 + 1)(n^5 - n), and n^5 - n == n - n == 0 mod 5. So 5 divides M.
(We could again show that 3 and 2 are divisors because M's is a multiple of (n^2 - n) and of (n^3 - n), and n^2 - n == n - n == 0 mod 2, and n^3 - n == 0 mod 3.)
Thus we have shown that 2, 3, 5 and 17 are always divisors of M.
And 257 is not always a divisor.
The GCD is 510.
well done, you just still need to justify why 257 is not always a divisor.
2:22 n^17 isn't n^4 mod 3.
Thank you for noticing that, it is typing mistake, it should be n^3 not n^4, final result is still the same.
Explain the problem more. We are looking for the greatest common divisor of n^17 - n and n? It would have to be n.
Are we looking for the greatest common divisor of n^17 and n? It would also be n.
Are we looking for the greatest number that is a gcd of n^17 - n and any real number? It would be n^17 - n.
Or maybe we are just looking for the greatest divisor of n^17 - n. No "common".
But the greatest divisor of n^17 - n would still be n. So we are really looking for the greatest number d that divides n^17 - n for any n.
The level of pendanticness reminds me of the good old days ; )
Yeah, that was my first reaction, what is "gcd" doing in the first statement of the problem????
It would be better stated as "find the largest k in N such that k is a divisor of n^17-n for all n in N"
why is'nt this channel blowing up
It's nonsingular. No blowup needed.
this is SUPERB
Thanks Tate, I'm literally 10 years too old for this stuff
To be fair, Andrew Tate sounds good. He would be considered to be an excellent success if only he took back all the thing he said about money, people, society, etc.
This will be my favorite channel
average trivial b orders 2 club question
Do Travis Bickle.
More of this for (Tingkatan 2)Malaysian?
You need to explain more.
this is what a sigma does for a living!
Andrew Tate is a legend 😂
Is this an IMO problem?
It is for preparing to the IMO but not an IMO problem itself. They are usually harder.
If you're going to do Tate, at least do a matrix problem.
More of Andrew Tate!!,mybe calculus or real analysis..#AikidoAnalysis
1.) We know it must be a product of primes to the first power. It cannot contain a square, because p^2 wouldn't divide p^17-p.
2.) It suffices to check for prime factors less than 17, because otherwise we know that there exists a primitive root of p, therefore an n such that p is the smallest number satisfying p | (n^p-n) therefore p doesn't divide n^17-n.
Thats cap bro
Is this voice simulations? Impersonation? Real Andrew? WTF>
It's AI generated
its the real andrew tate cant you hear his voice?
Even Biden could solve this problem.
Tate W
Didnt understand shit
'Promo sm'
Check Mr beast working on a math Olympiad problem :
ua-cam.com/video/ygl240Y3hjw/v-deo.html