13rd 1 can be done by induction,i guess
Hey guys , hope you are not too bored with maths . I found some good questions which I think don’t require pen and paper , yet are interesting . So please try ---
1 > If xy = 281 . 378 . 534 ( x + y ) , then find the number of integral solutions .
2 > If n = 640640640646 , then prove that n2 will leave a remainder 4 when divided by 8 .
3 > If p is a prime number , then prove that 1p-1 + 2p-1 + 3 p-1 ...........( p -1 )p-1 + 1 = 0 ( mod p )
-
UP 0 DOWN 0 0 9
9 Answers
212)
640640640646≡-2mod8
squaring
640640640642≡4mod8
H.P. that n2 will leave a remainder 4 when divided by 8
1@Eragon yep you are absolutely correct @Sourav , try thinking about congruencies (though i am not saying that induction can't be done) First one is easy , why people are not trying it ?
well , I am posting the solutions in one hour anyway.
211p-1 + 2p-1+3p-1................(p-1)p-1+1
add and sub (p-1) and (p-1)
(1p-1 -1 ) + (2p-1 -1)........................((p-1)p-1 -1) +(p-1+1)
=(1p-1 -1 ) + (2p-1 -1)........................((p-1)p-1 -1) +p
now from fermets little theorem we know tat if a is an integer coprime to p and p is a prime
then ap-1 -1≡0mod p
(1p-1 -1 )≡0modp
(2p-1 -1)≡0modp
similarly doing it
and p≡0modp
hence 1p-1 + 2p-1+3p-1................(p-1)p-1+1≡0modp
1Good job really , Eragon . Solution for first one ----
Take 281 . 378 . 534 = a
so given eqn. xy = a(x+y)
or , xy - ax -ay =0
or . (x-a)(y-a) = a2 (juct factorising)
so (x-a)(y-a)=2162 . 3156. 5 68
now no. of integral solutions = no. of divisors of that filthy number = (163)(157)(69)
1can any one expalin sawmya's ans and eragon's too....btw wat is fermats theorem
39see this kaustab:
http://en.wikipedia.org/wiki/Fermat's_little_theorem
and to get a hold on what is congruencies, see this:
http://targetiit.com/iit-jee-forum/posts/congruences-a-z-11709.html