I think that it should rather be $(a+d) \equiv (a+b) \pmod{10} \text{ or } a+d \equiv a+b+1 \pmod{10}$Labib wrote:From the problem it's clear that the palindrome's first and last digit would be $(a+d)$ and $(a+b)$. now,
$(a+d)=(a+b) => b=d$
It can easily be seen that if $(a+d)=(a+b)\geq10$ the number $abcd+bcda$ wouldn't be a palindrome!!
Combinatorics Marathon!!
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: Combinatorics Marathon!!
Moon vai, I fixed the solution and hoping this time it's ok this time. please check???
btw please solve my prob, somebody!! is there any theo I can use for it???
btw please solve my prob, somebody!! is there any theo I can use for it???
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Re: Combinatorics Marathon!!
I'll probably check later, but I guess that it is correct (you have got the idea at least)...
You see I am also a 'muscle' of this forum...I need to do the coding/modification stuffs as well.
You see I am also a 'muscle' of this forum...I need to do the coding/modification stuffs as well.
"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.
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Re: Combinatorics Marathon!!
It's ok. I see you're working really hard!! 
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes