Divisibility 1,2,3,4,5,6,7,8,9,&10 Tried:
Seems the ten-digit number ends with $240$ or $640$ or $840$ (Is not true, there are more ways the number could end)

$8325971640,$
$8365971240,$
$8317956240,$
$8291357640,$
$8325971640,$
$8235971640,$
$1357689240,$
$1283579640,$
$1783659240,$
$1563729840,$
$1763529840,$
$1653729840,$
$7165239840,$
$7195236840,$
$2165937840,$
$9283579640$
Suppose the number is of form $N=jihgfedcba$. We may write:

$a+b+c+d+e+f+g+h+i+j+(10-1)b+(10^2-1)c+(10^3-1)d+(10^4-1)e+(10^5-1)f+(10^6-1)g+(10^7-1)h+(10^8-1)i+(10^9-1)j$

A: Any number such as following forms are divisible by 2, 4, 5 and 8:

$(2k)(40)$ such as  $240, 440, 640, 840 . . .$

$(2k+1)(20)$, such as $120, 320, 520, 720, . . .$

B: Whatever the value of g is, the term $\frac{10^6-1}{9}$ is divisible by $77$.

C: For 7 we consider the remainder of $10^n-1$ when divided by 7:

$T=.....10, 10^2, 10^3, 10^4, 10^5, 10^6, 10^7, 10^8, 10^9$

$R_{10^n}=...3,..2,.. 6,.. 4,.. 5,... 1,.. 3,.. 2,.. 6$

$R_{10^n-1}=2,..1,..5,..3,...4,...0,...2,..1,..5$

We can make following relation for divisibility by 7:

$a+b+c+d+e+f+g+h+i+j+(2)b+(1)c+(5)d+(3)e+(4)f+(0)g+(2)h+(1)i+(5)j≡ mod 7$

D: For 11 we just consider the remainder of $10^n-1$ for odd n because for even n,. $(10^n-1)$ is divisible by 11  :

$T= .......10,....10^3,...10^5,...10^7,..10^9$

$R_{10^n}=...-1,...-1,...-1,...-1,..-1$

$R_{10^n-1}=-2,...-2,...-2,...-2,..-2$

So we can make following relation for divisibility by 11:

$a+b+c+d+e+f+g+h+i+j+(-2)b+(-2)d+(-2)f+(-2)h+(-2)j≡ mod 11$

So we have following system of Diophantine equations:

$a+b+c+d+e+f+g+h+i+j+(-2)b+(-2)d+(-2)f+(-2)h+(-2)j≡ mod 11$

$a+b+c+d+e+f+g+h+i+j+(2)b+(1)c+(5)d+(3)e+(4)f+(0)g+(2)h+(1)i+(5)j≡ mod 7$

The sum $a+b+c+. . .i+j= \frac{9(9+1)}{2}=45$ is divisible by 3 and 9. This system of equations indicates that the question can have numerous solutions, to find one for example take $cba=840$ which is divisible by 2, 3, 4, 5, 7 and 8, That is we assume $a=0$, $b=4$ and $c=8$ and look for other digits as follows, we have:

$45+4\times2+8\times1+5d+3e+4f+2h+i+5j≡ mod 7$

Or:

$61+5d+3e+4f+2h+i+5j≡ mod 7$

$45-2\times 4-2(d+f+h+j)=37-2(d+f+h+j)≡ mod 11$

Suppose $37-2(d+f+h+j)=11$ ⇒$d+f+h+j=(37-11)/2=13$

Suppose $d=1, . f=2,.h=3,. and,..j=7$ then we have:

$61+5+3e+8+6+i+35=115+3e+i≡ mod 7$

Let $115+3e+i=21\times 7=$⇒ $3e+i=32$ ⇒ $e=9$ and $i=5$

The only number which remains is 6 for g, so one solution can be:

$N=7536291840$
The number which are divisible by $8$ is also divisible by $2$ and $4$.

The number which are divisible by $9$ is also divisible by $3$.

The number of which is divisible by $6$ is also divisible by $2$ and $3$.

The number which are divisible by $10$ is also divisible by $2$ and $5$.

Also also the number we expect is divisible by $11$ and $7$.

So the number is in the form $=P×2^{3i}×3^{2j}×5^k×7^m×11^n$, 
Where $i$, $j$, $k$, $m$, &amp; $n$ are positive any positive integer and $P$ is any positive integer integer.Using this condition we will produce a required ten digit number.