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This variety of question is sure to not to be asked in JEE-Main. What do you think?? (+1)
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No , but it's under b i n o m i a l t h e o r a m .
η = 2 2 2 2 5 5 5 5 = ( 3 1 7 × 7 + 3 ) 5 5 5 5 .
Let us define R n ( . ) as remainder when ( . ) by n .
R 7 ( η ) = 3 5 5 5 5 = 2 4 3 1 1 1 1 = 5 × 5 1 1 1 0 = 5 × ( − 1 ) 3 7 0 = 5
I also believe that question level in this set is much higher than mains level. Many questions have more than 4 options while in jee mains, all the questions have 4 options.
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I also feel the same that's why I would not be attempting this set any more.. Questions are good but are irrelevant for JEE mains..
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@Rishabh Jain – Come on slack! Exams over?
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@Akshat Sharda – He has JEE...
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@A Former Brilliant Member – Oh! I just forgot about it!
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@Akshat Sharda – Yeah.. Surely I'll join after JEE ADV in June... :-)
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@Rishabh Jain – soo, what rank did you get in jee adv 2016 ?
Yeah NT is not there for JEE I think isnt it ?
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Exactly.. JEE Main questions are straightforward formula putting( with few exceptions)... And NT is not in syllabus.. Even in ADVANCE NT is not in syllabus..
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@Rishabh Jain – Yeah they are in Olympiad syllabus but I being in 10th solve all types of questions.
@Rishabh Jain – I think you are right!
@Rishabh Jain – So hows your preparation going :) ?
BTW, this question is in the example section of arihant publication book used for jee (author:- S.K Goyal)..
What is $\phi$ here
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2 2 2 2 5 5 5 5 ≡ 3 5 5 5 5 ( m o d 7 ) 3 ϕ ( 7 ) = 3 6 ≡ 1 ( m o d 7 ) 3 5 5 5 5 ≡ 3 5 ≡ 5 ( m o d 7 )