## BdMO 2017 junior/4

- samiul_samin
**Posts:**1007**Joined:**Sat Dec 09, 2017 1:32 pm

### Re: BdMO 2017 junior/4

Another solution is here.

- samiul_samin
**Posts:**1007**Joined:**Sat Dec 09, 2017 1:32 pm

### Re: BdMO 2017 junior/4

Just use this sign($) at the star and end of every equation.It will make your post more readable.prottoydas wrote: ↑Sat Mar 03, 2018 4:34 pm$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 like that

Code: Select all

`$2^2+3^n=7^1$`

Learn

**LaTeX**here.