Here's the question you clicked on:
shubhamsrg
if [x] denotes greatest integer < or = x , then prove that C(n,3) - [ n/3 ] is a natural no. , divisible by 3 for all integer n> or =3
C(n,3) here denotes comination, for e.g. C(4,3) = 4! / (4-3)! (3)! likewise..
|dw:1345190132115:dw||dw:1345190209186:dw| n can have only 3 from ... 1) 3k 2) 3k+1 3) 3k+2
am all years..what next ?
|dw:1345190372362:dw| this is just a try man ... i never know the solution beforehand.
that 4 wont be there in the last step...anyways nice... it'll be always divisible by sine its 9*something..and one of k and k-1 is always even so the 2 cancelled out..thus 3k satisfies it.. nice..
i meant it'll always be divisible by 3**
lol ... how ... you always need k of the form to be k, k - 1, k - 2
Oh ... sorry ... such a huge error!! :(((
same thing repeats in other cases .. since you have three terms of m, m+1, m+2 <--- one of them is going to be divisible by 3
and that [n/3] thing is not going to change.
excellent,,thans a lot :)
not really ... i guess i should get some more sleep :)
yeah you must be tired..go and relax..even very sharp minds need sound sleep! ;)