Consider $T$ = $9 \times 99 \times 999 \times 9999 \times \cdots \times \underbrace{999....9}_{2015 \:nines}$ Consider $T$ = $9  \times 99 \times 999 \times 9999 \times 
 \cdots \times \underbrace{999....9}_{2015 
\:nines}$ 
Find the last 3 digits of $T$
Advise: I wrote it down wrong the first time, it should be a product of "2015" numbers, i apologize about that, i realized my fault when i was travelling and i couldn't repare it in my cellphone.
My try
I know the last digit, i found it easily, but the struggle is with the others. I tried this:
$9  \times 99 \times 999 \times \cdots \times \underbrace{999 \ldots 9}_{2015 
\:nines}$  $= 9 \times 9(11) \times 9(111) \times \cdots \times 9(\underbrace{111 \ldots 1}_{2015 
\:ones})$
So $T$ = $9(1+11+111+ \cdots +\underbrace{111 \ldots 1}_{2015 
\:ones})$
But from here i found nothing, any hints?
 A: Hint:
$$T\equiv9\cdot99\cdot999^{2015-2}\pmod{1000}$$
Now $999\equiv-1\pmod{1000}\implies999^{2013}\equiv(-1)^{2013}\equiv-1$
$$\implies T\equiv(10-1)(100-1)(-1)\equiv-1+10+100$$
A: The question was changed from a product of four to a product of 2015 factors.
Maybe the author of the question may change the number of factors again, so let's solve it for any number of factors $n \geq 3$:
$$\prod_{k=1}^n\sum_{i=0}^{k-1}9 \times 10^i \equiv 9 \times 99 \times \prod_{k=3}^n999 \equiv 9\times 99 \times (-1)^{n-2} \equiv \left\{ \begin{array}{ll} 109 \mbox{ mod } 1000 & n= 2k+1 \\ 891 \mbox{ mod } 1000 & n = 2k \end{array} \right.$$
So, for $n = 2015$ we get $109$.
I leave the cases $n=1$ and $n=2$ for the inclined reader :-).
A: $T=891\times (1000a_3-1)(1000a_4-1)(1000a_5-1)\cdots(1000a_{2015}-1)$ for some integers $a_3,a_4,a_5,\dots,a_{2015}$.
So, $T=891\times (1000a-1)$ for some integer $a$.
The last three digits are $109$.
A: Well, following from where you ended up:
$T = 9(1 \times 11 \times 111 \times ...)$
Notice that except for the first and second terms, the other 2013 terms are congruent to 111 (mod 1000).
So $T \equiv 9(1 \times 11\times 111^{2013}) \pmod {1000}$
We will use the property that $\varphi (n^m) = n^{m-1} \varphi (n)$
So $\varphi (10^3) = 10^2 \varphi (10) = 100 \times 4 = 400 $
Since $2013 = 400 \times 5 + 13$,
$9(1 \times 11\times 111^{2013} \equiv 9(11 \times 111^{13}) \equiv 9(11 \times 111^{10} \times 111^3) \equiv 9(11 \times (111^2 )^5 \times 1367631) \equiv 9(11 \times 12321^5 \times 631) \equiv 9(11 \times 321^5 \times 631) \equiv 9(11 \times 3408200705601 \times 631) \equiv 9(11 \times 601 \times 631) \equiv 9(4171541) \equiv 9(541) \equiv 4869 \equiv 869 \pmod {1000}$
