623> Prove that n ( n6 - 1 ) is always divisible by 42.
this is same as prove that n7-n is always divisible by 42
by fermat's, we can prove that n7-n is always a multiple of 7
also we can say that the number is even because either n or n6-1 is even....
also we can say that n(n6-1) is a multiple of n(n2-1)
applying fermat's again we can say taht n3-n is a multiple of 3
hence the given number is a multiple of 7, 3 and 2 .. hence a multiple of 42
62
Lokesh Verma
·2009-11-16 02:40:06
Read these posts for coefficient of x^k in the expansion type of problems...
http://targetiit.com/iit-jee-forum/posts/binomial-dbt2-11998.html
http://targetiit.com/iit-jee-forum/posts/summation-903.html
http://targetiit.com/iit-jee-forum/posts/eazzy-9162.html
http://targetiit.com/iit-jee-forum/posts/binomial-1s-10954.html
Read the proof of wilson's here: http://en.wikipedia.org/wiki/Wilson%27s_theorem
I am too lazy to type the proof :P
1Absolutely correct sir, please let me give the solution for the first one ----
79 + 97 is an even number , so p can take values of 62 , 64 ,66 , 68 only.
Now 7 ≡ -1 (mod 8)
or, 79 ≡ -1 (mod 8)
so 79 + 1 is divisible by 8.
Again , 9 ≡ 1 (mod 8)
so 97 ≡ 1 (mod 8)
i.e, 97 - 1 is divisible by 8 .
so (97 - 1) + (79 +1) is divisible by 8, or 97 + 79 is divisible by 8.
So p should a multiple of 8.
So p = 64
Now x3 + px -5p = 0 becomes
x3 + 64x - 5 * 64 =0
or , x = 3√64( 5 - x )
clearly , x = 4 is the only real solution.
341even numbers can be multiples of odd numbers. so eliminating the odd numbers there isnt right
1
Maths Musing
·2009-11-18 10:21:38
Thank you sir for completing my solution --- really I hadn't thought about it.
