1no this isnt my doubt[1]
let a1,a2,a3.......an be n nos such that each a2 is either 1 or -1....If a1a2a3a4 + a2a3a4a5+................ana1a2a3=0.............then prove 4 divides n[1]
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14 Answers
62is this a doubt?
This one is from Arthur if i remember correctly!
62hmm.. ISI mocks have become far far tougher than they used to be at my time..
i remember i had almost solved the whole paper in their entrance exam...
but with these questions i wonder how many jee aspirants are able to solve even half of it!
1mm..okh then leave it.....i guess it may be a mere time wastage on our part to be solving such kinda q's....as they wont help in jee
btw i dont know of the level of isi nowadays.....is it this tough
62hmm... Riemann it is difficult but not impossible..
It uses a very nice logic ..
lets see the number of terms where each ai appears... (4)
they are 4 ..
now the possibilities are that there could be
all these 4 terms as 1
or
3 as 1 and 1 as -1,
or
2 as 1 and 2 as -1
or
1 as 1 and 3 as -1
or
no 1 while all 4 -1
now in each of these cases, observe that if you replace ai by -ai, the sum will change in the cases above by
8, 4, 0, 4 and 8 respectively
so the sum remains unchanged modulo 4 (remainder on division by 4)
After this observation, what we can do is replace all the -1's to 1s
which means that the remainder on division by 4 remains same..
but that would also mean that the sum we have becomes 1+1+1... n times = n
hence n is divisible by 4
[1]
62This one is probably the same thing from Arthur angel and not my solution.. i had read it a few months back..