Imo 2007-5
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- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
then d is divisor of $4ab-4a^2$ and if d don't accept 4 as a value then d is factor of a. my request try to understand that then u find a contradiction.
Re: Imo 2007-5
Uhh. What about $d=7,a=3,b=10$?
Then $7|4*10*3-1$ and $7|4*3^2-1$ and $7|4*10^2-1$ and $7|10-3=a-b$ and $7$ does not divide $10$ or $3$. So, I can't find any contradiction!
Then $7|4*10*3-1$ and $7|4*3^2-1$ and $7|4*10^2-1$ and $7|10-3=a-b$ and $7$ does not divide $10$ or $3$. So, I can't find any contradiction!
One one thing is neutral in the universe, that is $0$.
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- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
Re: Imo 2007-5
sadly u r right! but i find my mistake, $d$ divisor of $4a^2-1,4ab-1$ and $4ab-4a^2$,it doesn't means $d$ factor of both $4ab$ and $4a^2$ but $d$ divisor of $4ab-1$ and $4a^2-1$ actually $d$ factor of $4a(b-a)$ then since $4a$ not divide $A,B$ so $d=(b-a)$ [here b>a]. thanks for ur kind reply but u can feel happy for that at least u make someone understand.
$d=(b-a)$,i will try to prove it from here.thank u.
$d=(b-a)$,i will try to prove it from here.thank u.
Re: Imo 2007-5
Still that's wrong. This doesn't mean that $d=a-b$, only $d|a-b$.tarek like math wrote:sadly u r right! but i find my mistake, $d$ divisor of $4a^2-1,4ab-1$ and $4ab-4a^2$,it doesn't means $d$ factor of both $4ab$ and $4a^2$ but $d$ divisor of $4ab-1$ and $4a^2-1$ actually $d$ factor of $4a(b-a)$ then since $4a$ not divide $A,B$ so $d=(b-a)$ [here b>a]. thanks for ur kind reply but u can feel happy for that at least u make someone understand.
$d=(b-a)$,i will try to prove it from here.thank u.
One one thing is neutral in the universe, that is $0$.
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- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
Re: Imo 2007-5
oh, it's a mistake but in previous line i say $d$ factor of $4a(b-a)$ then i missed it.
Re: Imo 2007-5
But I think it would be a better suggestion for you that don't try a problem from only one fixed view. Because your approach might be wrong.
One one thing is neutral in the universe, that is $0$.