BdMO 2017 junior/4

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samiul_samin
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Re: BdMO 2017 junior/4

Unread post by samiul_samin » Sat Mar 03, 2018 7:39 pm

Another solution is here.

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samiul_samin
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Re: BdMO 2017 junior/4

Unread post by samiul_samin » Sun Mar 04, 2018 5:12 pm

prottoydas wrote:
Sat Mar 03, 2018 4:34 pm
$n^3-n^2=n^2(n-1)$
Here we have 2 cases.
Case $1$:
$n^2$ is divisible by $73$ and $n-1$ is divisible by $27$.
Now $n$ is the form of $73k$.
We get $73k-1$ is congruent to $0$(mod$27$)
Or,$-(8k+1)$ is congruent to $0$(mod$27$)
Or, $8k$ is congruent to $26$ (mod$27$)
Or, $4k$ is congruent to $13$ (mod$27$)
Or, $4k$ is congruent to $40$ (mod$27$)
Or, $k$ is congruent to $10$(mod$27$)
So, $k=27m+10$ (where $m=0, 1, 2, 3, ………$).Taking the smallest value of $m=0$ gives $k=10$ and $n=73\times 10=730$.
Case$2$:
$n^2$ is divisible by $27$ and $n-1$ is divisible by $73$.
So, $k$ is the form of $73K+1$.
We get $(73K+1)^2$ is congruent to $0$(mod$27$)
Or $5329K^2+146K+1$ is congruent to $0$(mod$27$)
Or, $10 K^2+11K+1$ is congruent to $0$(mod$27$)
Middle term of it gives,$(K+1)(10K+1)$ is congruent to $0$(mod$27$)
If we take $K+1$ is congruent to $0$(mod$27$)
We get the smallest value of K is 26.
Now
$10K+1$ is congruent to $0$(mod$27$)
Or, $10K$ is congruent to $26$(mod$27$)
Or, $5K$ is congruent to $13$(mod$27$)
Or, $5K$ is congruent to $40$(mod$2$7)
Or, $K$ is congruent to $8$(mod$27$)
So, $K=27M+8$(where $M=0,1,2,3,………$).Taking the smallest value of $M=0$ gives $k=8$ and $n=73*8+1=585$
Just use this sign($) at the star and end of every equation.It will make your post more readable.
Just like that

Code: Select all

$2^2+3^n=7^1$
will become $2^2+3^n=7^1$
LearnLaTeX here.

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