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7 2 1 + 7 2 2 + 7 2 3 + 7 2 4 is divided by 25. find the remainder
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2 solutions
Just 7 2 ≡ − 1 ( m o d 2 5 ) should suffice. Don't need to calculate 1 + 7 + 7 2 + 7 3 = 4 0 0 .
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Good point. I just found it a nice result that those four powers of 7 add to 400. :)
Corollary:
∀ m , n ∈ Z 0 + ( n > m ) , n − m ≡ 3 ( m o d 4 ) ⟺ k = m ∑ n 7 k ≡ 0 ( m o d 2 5 )
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Cool. Have you ever thought of becoming a mathematician, Prasun? I'm serious, (for once). :)
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Nope, you're still joking! :P
Seriously though, this comment, coming from you means a lot to me. And the answer to your question is indeed yes since my dream is to go into research in pure mathematics. But then again, the obstacle is the education system here in India which solely focuses on your marks in exams rather than your understanding of the subject. And most of the good colleges here have so high cut-off marks that I'm being told by some that I might not get a chance to pursue higher mathematics because of my low 82% marks in finals. :(
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@Prasun Biswas – Yeah, marks and understanding don't always correlate. From what I've seen here on Brilliant, your understanding is undeniable; the field of mathematics would benefit from you, I'm sure. I don't know if there's a chance that you could study in North America; marks are still important to get in, but they tend to take a broader look at applicants, reasoning that marks never tell the whole story about the potential of an individual. I hope that you're able to find some way to achieve your dream. Just never take "No" as an answer, (and I know I'm going to get a clever reply to this last comment). :D
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@Brian Charlesworth – I'd love to get a chance to study higher mathematics abroad (not necessarily America) but that's probably not going to happen considering my marks and financial conditions.
Nonetheless, thank you for your kind words. It sure did make my day. :)
Just letting you know, I'm up for employment if someone needs an assistant (without degree :P) for mathematical work (research maybe). So, if you ever need an assistant for math-related stuff, you can hire one by contacting our assistant-providing service at the following e-mail address:
prasunbiswas.2012@gmail.com
:P
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@Prasun Biswas – Hahaha. I'll definitely keep you in mind. :)
thanks for the solution.... i wanted to know Mr. @Brian Charlesworth , that do you get points to post a question or to post an answer to a question? i want to know!
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There are no points given for posting questions or solutions; doing so is its own reward. :)
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and who are challenge masters?
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@Sarthak Rath – I think that Calvin Lin is the primary Challenge Master, but there may be others.
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@Brian Charlesworth – you? or u know? i guess u are as ur levels are pretty awesome!!!
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@Sarthak Rath – Haha. Thanks, but no, not me. I'm just a regular member. :)
Would it be alright to say that since 4 0 0 is a multiple of 1 6 , the remainder will always be 0?
Take 7 2 1 common 7 2 1 ( 1 + 7 1 + 7 2 + 7 3 ) = 7 2 1 ( 4 0 0 ) Which is divisible by 2 5
yarr... tu jab 2 digit ka power chadha ta hai, to power pe {} laga like in this case {21} or your latex will be read as it is now... pl.rectify!
this is exactly my approach i told u in class... also this is brian charlesworth's approach! cheater!!! (no offence :::....)
7 2 1 + 7 2 2 + 7 2 3 + 7 2 4 = 7 2 1 ( 1 + 7 + 7 2 + 7 3 ) = 7 2 1 ∗ 4 0 0 =
7 2 1 ∗ 1 6 ∗ 2 5 ≡ 0 ( m o d 2 5 ) .