Even,You are odd looking!!!!!!!!!

[sakibtanvir]

Consider a $17$ digit number $N=10^{18}x_0+10^{17}x_1+10^{16}x_2+........10x_{17}+x_{18}$.Another number $M=10^{18}x_{18}+10^{17}x_{17}+...+10x_1+x_0$.Prove that,at least one digit of $(M+N)$ is even.

[sakibtanvir]

Hey,No one interested ???

[sm.joty]

Hey,No one interested ???

That's very easy one so may be no one want to make a reply. Anyway there is a bug in your problem. You said it is a 17 digit number but your $N$ and $M$ representation says they are 19 digit number

ok it's not a matter you can also solve it. here is the solve but now generalize it for All odd digit numbers.

just sum up and you'll get $10^{9}.2.x_9$
that line contains your answer.

[sakibtanvir]

You have done the same mistake as mine.It would be $10^{8}2x_{9}$

[sakibtanvir]

OOps,not just this part,I think ur solution is totally wrong.Because,After summing, we get,\[10^{16}(x_{1}+x_{17})+10^{15}(x_{2}+x_{16})+10^{14}(x_{3}+x_{15})+...\]
Look that,\[(x_{1}+x_{17}),(x_{2}+x_{16}),(x_{3}+x_{15})+...\] can be greater than $9$.So they are not digits(1,2,......9,0) so 1 can be added with $10^{8}2x_{9}$ that makes it odd.

[SANZEED]

How can you tell that they won't be $\geq 9$?

[sakibtanvir]

Who told it would not be greater than 9?????????

[sm.joty]

OOps,not just this part,I think ur solution is totally wrong.Because,After summing, we get,\[10^{16}(x_{1}+x_{17})+10^{15}(x_{2}+x_{16})+10^{14}(x_{3}+x_{15})+...\]
Look that,\[(x_{1}+x_{17}),(x_{2}+x_{16}),(x_{3}+x_{15})+...\] can be greater than $9$.So they are not digits(1,2,......9,0) so 1 can be added with $10^{8}2x_{9}$ that makes it odd.

Well I don't understand why you trying to make me wrong
Lets see,
Suppose we have 19 boxes, and they are
0,1,2,3,4,5,6,7,8,9,10,11,12,14,15,16,17,18
18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0
now make pairs like
$(0,18),(1,17),(2,16),(3,15),(4,14),(5,13),(6,12),(7,11),(8,10),(9,9),...............(18,0)$
(A painful work for your understand )
see the only pair where there is two repeated is (9,9).
Now I think you are clear about the exponent of ten.
well now back my solution.
see the number $10^{9}.2.x_9$ have 9 trailing zeros.let $2.x_9=ab$ [where $pq$ is not $p*q$, here $pq$ is like $13,p=1,q=3$]
The next term $10^{10}.(x_{10}+x_8)$ have 10 trailing zeros.let $10^{10}.(x_{10}+x_8)=ab0000000000$
where ab is not $a*b$, here $ab$ is like $13,a=1,b=3$
now sum up tem
ab0000000000
pq000000000
=a(b+p)q000000000
So you understand that here we need to consider only unit digit.The other digit will be added with the next term's unit digit.
now see all $2.x_9$ numbers are even and there unit digit is also even.
It makes our answer.
Any doubt ?

[sakibtanvir]

I thought u have done this with 17 digits...But i did not understood u have worked out with 19 digits.For 19 digit ur exponent it right...Now come to the other.........................................................
U have misundertood me...u thought that, I claimed ur summation wrong...No,ur adding is absolutely right.I have not argued about this..I just said that, your logic does not conclude the solution.U have claimed that,The 10th digit will be even.I want to know how can u be so sure about that????? After summing It can be odd also.(If u can not fix a digit as even, solution has no value.) Hope u have got what i am wanting to say....I think others will agree with me..(I can be wrong also)