BdMO Online Forum

Posted: **Sun Dec 12, 2010 9:07 pm**

O... hum, making too many silly mistakes

Posted: **Sun Dec 12, 2010 10:05 pm**

^Don't become a "Moon"

Posted: **Sun Dec 12, 2010 11:38 pm**

Actually it is not any mistake,just trivial case

.Nothing to be sad at all,I don't understand why you are sad.

Posted: **Mon Dec 13, 2010 5:19 am**

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

Posted: **Mon Dec 13, 2010 10:10 pm**

I think Avik Vaia has understood by this time why I mentioned this as the easiest problem of APMO

Posted: **Tue Dec 14, 2010 12:52 am**

I still don't prefer using difficulty denoting adjectives with problems...

Posted: **Mon Jun 27, 2011 3:41 pm**

' Oshim obonmon ' gives a short solution ......

Posted: **Mon Jun 27, 2011 3:46 pm**

Ops .. I didn't noticed that ovik da has given the solution ..... (by this method )

Posted: **Mon Jun 27, 2011 5:54 pm**

Nadim Ul Abrar wrote:Ops .. I didn't noticed that ovik da has given the solution ..... (by this method )

He didn't use Infinite DEscent.

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.

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!

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