User:Idk12345678

Revision as of 23:00, 6 June 2024 by Idk12345678 (talk | contribs)

My Solutions

2001 AMC 10 Problem 10

2001 AMC 10 Problem 12

2002 AIME II Problem 3

2005 AIME II Problem 5

2008 AMC 10B Problem 3

2008 AMC 10B Problem 6

2008 AMC 10B Problem 9

2008 AIME II Problem 1

2009 AMC 10A Problem 10

2009 AIME II Problem 2

2013 AMC 10A Problem 16

2014 AMC 10B Problem 9

2018 AMC 10A Problem 10

2020 AMC 10A Problem 14

2023 AIME I Problem 2

2024 AIME I Problem 2

Some Proofs I wrote

$(x+y)^n \equiv x^n + y^n \pmod{n}$ if $n$ is prime.

Proof: Expanding $(x+y)^n$ out, all the coefficients are of the form $n \choose r$ by the binomial theorem. To prove the original result we must show that if $r \neq 1$ and $r \neq n$, then \[{n \choose r} \equiv 0 \pmod{n}\]. Because \[{n \choose r} = \frac{n!}{r!(n-r)!}\], \[{n \choose r} \times r!(n-r)! = n!\], which is divisible by $n$, so the original expression must be divisible by $n$. However if $n$ is prime, \[\gcd(n, r!(n-r)!) = 1\], since $r!$ does not contain $n$(because $r<n$). Therefore, in order for \[{n \choose r} \times r!(n-r)!\] to be divisible by $n$, $n \choose r$ is divisible by $n$. All the coefficients of the expansion(besides the coefficients of $x^n$ and $y^n$) are of the form $n \choose r$, and \[{n \choose r} \equiv 0 \pmod{n}\], so they cancel out and \[(x+y)^n \equiv x^n + y^n \pmod{n}\] if $n$ is prime. $\square$

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