Such Big, Much Scare, Many Easy!

By multiplying the 1st 7 digits by 8 we get the last 9 digits as result.

$\Rightarrow$ $1002004$ is a factor of the $1002004008016032$

And, $1002004 = 2\times 2\times 250501$

Given that $P>250000$ , And P is Prime,

$\boxed{250501}$ is the answer.

$Split\quad it\quad up\quad and\quad write\quad neatly\quad to\quad get:$

$1002004008016032\ =\ { 2 }^{ 0 }*{ 10 }^{ 15 }+{ 2 }^{ 1 }*{ 10 }^{ 12 }+{ 2 }^{ 2 }*{ 10 }^{ 9 }+{ 2 }^{ 3 }*10^{ 6 }+{ 2 }^{ 4 }*{ 10 }^{ 3 }+{ 2 }^{ 5 }$

$Also\quad notice\quad that$

$\frac { { (x }^{ 6 }-{ y }^{ 6 }) }{ (x-y) } =(x+y)({ x }^{ 2 }+xy+{ y }^{ 2 })({ x }^{ 2 }-xy+{ y }^{ 2 })$

$={ x }^{ 5 }+{ yx }^{ 4 }+{ { y }^{ 2 }x }^{ 3 }+{ y }^{ 3 }{ x }^{ 2 }+{ y }^{ 4 }x+{ y }^{ 5 }$

$Now\quad let\quad x={ 10 }^{ 3 }\quad and\quad y=2\\ So,$

$1002004008016032=1002*998004*1002004=1002*4*249501*4*250501$

$As\quad by\quad the\quad question\quad there\quad is\quad a\quad prime\quad P>250000.$

$So\quad without\quad further\quad proving\quad that\quad 250501\quad is\quad a\quad prime,$

$we\quad can\quad assume\quad that\quad \boxed { P=250501 }$

It is an extremely easy problem. Took only a few seconds to solve. It should have been a level 2 problem.

Combining elements of both Karthik Sharma 's and Joseph Varghese 's solutions:

We first note that $3, 7$ divides our number, which will be important later.

We now consider making use of factorisation below:

$1000^5+ 1000^4*2^1+\cdots +1000^1*2^4+2^5=\frac{1000^6-2^6}{1000-2}$

$=(1000+2)(1000^2+1000*2+2^2)(1000^2-1000*2+2^2)$

$=1002*998004*1002004$

And we note that $1002=2*3*167$ , $998004=4*3*7*11881$ , since as we've observed since $3,7$ do not divide $1002004$ , they must divide some of the other factors.

$1002004=4*250501$ , which so far is the only non-factored factor of out number that we have so far that is $>250000$

(Side note: $11881$ further factorises to $109^2$ , but that's unimportant as it is already $<250000$ )

Hence, our desired factor is thus $\boxed{250501}$