2 0 2 0 2 0 2 8 m o d 2 0 2 7 = ?
All of my problems are original
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hey solve this problem @Vinayak Srivastava , @Aryan Sanghi , @Mahdi Raza Khunt .. https://brilliant.org/problems/wait-wait-wait-wait-think-if-you-can-designed-for/?ref_id=1599188 you can also look at it on my profile
@Aryan Sanghi post the solution
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let me see how your solution looks like
It's long, I'll post it when I'll get time. I did construction and completed the square of hexagon. I have seen this problem before, my sir gave me.
@in which grade are you
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In class 11th.
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I,m in class 7th
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@Srijan Singh – Ohk. That's nice. You posted good question. Keep it up.👍
@Srijan Singh – @Aryan Sanghi , I SOLVED AT THIS AGE
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@Srijan Singh – WE BOTH ARE IN FIITJEE YAY
@Srijan Singh – Hmmm. That's nice. You're doing nice. Keep it up.
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@Aryan Sanghi – @Aryan Sanghi ,THANKS FOR YOUR AASHIRVAD
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@Srijan Singh – Oh Lol. But, I'll advice you not to use capitals. It is considered shouting in internet. Maybe you don't know it. :)
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@Aryan Sanghi – okay thanks , i was unknown about this
@Aryan Sanghi in which grade are y ou
2 0 2 0 2 0 2 8 m o d 2 0 2 7 written this way, it means the remainder when 2 0 2 0 2 0 2 8 is divided by 2 0 2 7 , which is always non-negative.
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Sir, can I please make it look big again? It looks more clear in these short questions. :) @Chew-Seong Cheong
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Just use maximum \large will do. Don't make it like a kindergarten book, the larger the font the better.
@Aryan Sanghi .hey can you explain modular arithmetic to me and can you post a new question for everyone
@Aryan Sanghi look at my solution of problem is for testing skills
2 0 2 0 2 0 2 8 ≡ ( 2 0 2 7 − 7 ) 2 0 2 8 (mod 2027) ≡ ( − 7 ) 2 0 2 8 (mod 2027) ≡ 7 2 0 2 8 m o d ϕ ( 2 0 2 7 ) (mod 2027) ≡ 7 2 0 2 8 m o d 2 0 2 6 (mod 2027) ≡ 7 2 ≡ 4 9 (mod 2027) Since g cd ( 7 , 2 0 2 7 ) = 1 , Euler’s theorem applies. And the Euler’s totient function ϕ ( 2 0 2 7 ) − 2 0 2 7 − 1 , since 2027 is a prime.
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Excellent solution sir. Thanku for sharing it with us.
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By Fermat's Little Theorem , for any positive integer a and prime p
a p − 1 ≡ 1 m o d p
2 0 2 0 2 0 2 7 − 1 ≡ 1 m o d 2 0 2 7
2 0 2 0 2 0 2 6 ≡ 1 m o d 2 0 2 7 … ( 1 )
2 0 2 0 ≡ − 7 m o d 2 0 2 7
2 0 2 0 2 ≡ ( − 7 ) ( − 7 ) m o d 2 0 2 7
2 0 2 0 2 ≡ 4 9 m o d 2 0 2 7 … ( 2 )
2 0 2 0 2 0 2 6 × 2 0 2 0 2 ≡ 4 9 m o d 2 0 2 7 … (by (1) and (2))
2 0 2 0 2 0 2 8 ≡ 4 9 m o d 2 0 2 7