Easiest problem of Apmo
O... hum, making too many silly mistakes
Every logical solution to a problem has its own beauty.
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Re: Easiest problem of Apmo
^Don't become a "Moon"
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Re: Easiest problem of Apmo
Actually it is not any mistake,just trivial case .Nothing to be sad at all,I don't understand why you are sad.
One one thing is neutral in the universe, that is $0$.
Re: Easiest problem of Apmo
Ahh.... I wish I would become a "Moon"
Humm... sometimes these silly mistakes cause great losses. Moreover, I think doing silly mistakes means lack of attention
Humm... sometimes these silly mistakes cause great losses. Moreover, I think doing silly mistakes means lack of attention
Every logical solution to a problem has its own beauty.
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Re: Easiest problem of Apmo
I think Avik Vaia has understood by this time why I mentioned this as the easiest problem of APMO
One one thing is neutral in the universe, that is $0$.
Re: Easiest problem of Apmo
I still don't prefer using difficulty denoting adjectives with problems...
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
Re: Easiest problem of Apmo
Ops .. I didn't noticed that ovik da has given the solution ..... (by this method )
$\frac{1}{0}$
Re: Easiest problem of Apmo
He didn't use Infinite DEscent.Nadim Ul Abrar wrote:Ops .. I didn't noticed that ovik da has given the solution ..... (by this method )
Do you mean that $2^{p-q}-1=e.35$ instead of $f$? If so, then I see that $b=(d-e)2^{q-1},e\le d$, not leading to any contradiction!Avik Roy wrote:
Let,
$36a+b=2^p$
$a+36b=2^q$
Without loss of generalization, let $p>q$. Hence,
$37(a+b)=2^q(2^{p-q}+1)$
and $35(a-b)=2^q(2^{p-q}-1)$
From this we can deduce that,
$2^{p-q}+1=d.37$
and $2^{p-q}-1=\boxed f.35$
Thus $a+b=d.2^q$
and $a-b=e.2^q$
where $37d-35e=2$
It can be easily seen that $e \ge d$ and that leaves us with $b$ being negative.
One one thing is neutral in the universe, that is $0$.