Hunt For The Last 3

I don't get which series it was. Was it a series of odd numbers? then why 9 is not in the series and if it is the series of prime numbers then why 15 is in the series??

Can you hunt for the 9 digits in the image? Number 8 is white an in the lower right corner.

I just realised 0.o

When it says they want to find the last 3 digits, it means $\bmod {1000}$

You can read up what Chinese Remainder Theorem here

The prime factorization of $1000$ is $2^3 \times 5^3$ , with $2^3 = 8, 5^3 = 125$

Apply the properties of modular arithmetic:

$( x_1 \cdot x_2 \cdot x_3 \cdot \ldots \cdot x_n ) \bmod{M} \equiv \left ( (x_1 \bmod {M}) \cdot (x_2 \bmod {M}) \cdot (x_3 \bmod {M}) \cdot \ldots \cdot (x_n \bmod {M}) \right ) \bmod {M}$

Note that the remainder when divided by $8$ for $3,7,11,15,\ldots,2003$ yields $3,7,3,7, \ldots, 3$ respectively

Because $7 \equiv -1 \bmod{8}$ , we can rewrite them as $3,-1,3,-1, \ldots, 3$ . With $250$ of $-1$ 's and $251$ of $3$ 's. So the remainder of the expression when divided by $8$ is simply $(-1)^{250} \cdot 3^{251} \equiv 1 \cdot 3 \cdot 3^{250} \equiv 3 \cdot (3^2)^{125} \equiv 3 \cdot (9 \bmod 8)^{125} \equiv 3 \cdot 1 \equiv 3$

And it's explained above why the expression is divisible by $5^3=125$ , so the expression satisfy the linear congruences $x \equiv 3 \pmod{8}$ and $x \equiv 0 \pmod{125}$ , which gives the answer $875$