Why all odd numbers not ending with 5 divide exactly into a number comprising only 9's? Help me!!It's really frustrating I can't understand this simple thing.The maths instructor in my video,the renowned Arthur Benjamin,states (clip linked below) the following:
All odd numbers, that don't end in 5, divide exactly into a number that consists only of 9's.
http://www.youtube.com/watch?v=PufXAVtwmXk&feature=youtu.be
He takes 2009 as an example and goes on to explain it this way:
Take 2010(one more than 2009) numbers, with the first beginning with one 9,and the last comprising 2010 9's.
 9,99,999,9999,................9999999
                               ^ 2010 9's

Since there are 2010 numbers and 2009 possible remainders when divided by 2009,two of these numbers must have the same remainder.Let them be the following:
  999999...... i 9's= 2009p+r
  99999........j 9's=2009q+r

Subtracting both sides in above equations we get :
(number with i 9's - number with j 9's)= 2009(p-q)

So 2009 divides exactly into a number with i-j number of 9's followed by j 0's.
Now begins the part where I am so confused.At this point the instructor says that since 2009 is not a multiple of 2 or 5, we can remove the zeros from the left number.How is it justified? Please answer these three questions arising from this premise:
A) How can we remove the zeroes and expect the number to be still divisible by 2009?Like 200 is divisible by 25,but it's no longer divisible by 25 if we take out the last zero!! 40 is divisible by 8, but not if we take out the 0? Since Mr.Benjamin can't be wrong,can you tell me the reason why we can remove the zeroes?
B) Why can we remove the zeroes just because 2009 is not a multiple of 2 or 5?I know I am missing something simple here.....maybe I am having a mental burnout...so please answer this.
C) I tried to tinker with the instructor's explanation about how every odd number not ending with 5 divides into a number consisting of all 9's.And it seems ANY odd number not ending with 5 divides exactly into a number consisting only of ANY odd number, not essentially 9.Is it so?Is my assumption wrong?If so,why?Why can't we use the same logic and proof for numbers comprising only of 7's or 3's or 1's?
Please go easy on me this time as not only is this bottleneck from my very first video lecture on Discrete Mathematics,it is also my first question on Math.Stackexchange.Thank you!!
 A: It has to do with the particular form of your subtraction. Since you are subtracting a number with all $9$ from another with also all $9$, the result will be a number that has a number of $9$ on the left and then all the zeroes together on the right. 
The fact that $2009$ is not a multiple of $2$ nor $5$ tells you that all the powers of $10$ are on $p-q$, and so you can cancel them on both sides: on the left you'll get all the $9$, and on the right a multiple of $2009$. 
A: This is by a simple application of pigeon hole principle. Consider any odd number, say $a$, and look at the powers of $10$ mod $a$, i.e., $10^{n} \pmod a$. Now let $n$ vary from $1$ to $a+1$, then by pigeonhole principle, there are two numbers of the form $10^x$ and $10^y$ that leave the same remainder, where $1 \leq x < y \leq a+1$, i.e,
$$10^x \equiv r \pmod{a}$$
$$10^y \equiv r \pmod{a}$$
Subtracting the two we get that
$$10^y-10^x \equiv 0 \pmod{a} \implies 10^x \left(10^{y-x} - 1\right) \equiv 0 \pmod{a}$$
Since $a$ is odd, we have $\gcd(10^x,a) = 1$. Hence,
$$\left(10^{y-x} - 1\right) \equiv 0 \pmod{a}$$ and $10^{y-x}-1$ is a number of the form $\underbrace{99\ldots99}_{y-x \,\, 9's}$.
A: Using Carmichael function, $$a^{\lambda(n)}\equiv1\pmod n$$ if $(a,n)=1$
Any odd number $b$ is relative prime with $2$.
Any odd number $b$ is relative prime with $5$ if it does not end with $5$
So, $(b,10)=1\implies 10^{\lambda(n)}\equiv1\pmod b\implies b$ divides $(10^{\lambda(n)}-1)$ which consists of $\lambda(n)$ number of $9$s
