thx to the dude who provided the copy of the paper. kinda thrown off by this m/c a bit. Ended up picking C in the end just felt like it looks like a regular projectile motion graph.....probs got it wrong tho, anyone got any ideas?View attachment 33774
part i you can just do |xi +yj + zk|=1<=|xi|+|yj|+|zk| = |x|+|y|+|z| since you can multiply lengths of the unit vectors i j and k which are 1
Suppose T(2n) = H(m) for some integer m.How then?
85+What would a raw mark of 73 or 74 ish scale to roughly?
u contributed to the god-tier scaling, only thing u acedNah mate i reckon i'm gonna ace tf out the test
i did that in the examMeme Solution for 15. (d).
If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.
Hence, the result must be false for all n>2.
The result is clearly false for n=2 by manually checking.
QED
I posted a version of that in the discussion in the MX2 forum... I wonder how markers will evaluate anything like it?Meme Solution for 15. (d).
If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.
Hence, the result must be false for all n>2.
The result is clearly false for n=2 by manually checking.
QED
did you figure this one out? i also got lost with this lol and ran outta time to look at it properlyalso how did you do the triangle and hexagon numbers
i think a bit too much is sketched in there
I can't say. I know that for proving the irrationality of the nth root of 2, the proof of fermat's last theorem does in fact require some elementary irrationality results (including this very thing) so it would be circular reasoning. This would need the consultation of an expert in algebraic number theory to resolve.I WAS ABOUT TO DO THIS IS THIS THE CORRECT ANSWER??? IF IT IS IM GOING TO BE HANGING FROM THE CHANDELIER
specifying the branches of the function is a pain so i'm leaving that as an exercise to the readeri think a bit too much is sketched in there
i'm not the person u quoted and idk if this would be accepted but i noted that they want u to prove for every odd number and an odd number can be represented by 2n - 1 so....subbing 2n - 1 into the Triangle equation actually gets u the hexagonal equation, therefore hexagonal number??? seems way too simple lol (lmk if you think this would get any marks)did you figure this one out? i also got lost with this lol and ran outta time to look at it properly