Find $$ \prod_{i=1}^{1903} (2^i + 5) \mod 1000 $$
my try
I tried to find remainder $\mod 8$ and $\mod 125$ and use Chinese remainder theorem.
Mod 8
Let see that for each $2^k + 5$ where $k\ge 3$ we have $2^k + 5 \equiv 5$. So $$ \prod_{i=1}^{1903} (2^i + 5) \equiv 7 \cdot 9 \cdot 5^{1901} \equiv 35 \equiv 3 $$
Mod 125
Unfortunately I stucked at calculating mod $8$. I want to say that officially (that means: from lecture) I don't know Carmichael function but I know Euler function if it can be helpful there.
$\bmod125:$
By Euler's totient theorem we have $2^{100}\equiv1$. Factoring 100, we may also notice that $2^{50}=4^{25}\equiv-1$, which let's us reduce the problem down to solving
\begin{align}\prod_{i=1}^{100}(2^i+5)&\equiv\prod_{i=1}^{50}(25-4^i)\tag1\\&\equiv\prod_{i=1}^{25}(15625-16^i)\tag2\\&\equiv-\prod_{i=1}^{25}16^i\tag3\\&=-16^{325}\tag4\\&\equiv-1\tag5\end{align}
And so we have $\displaystyle\prod_{i=1}^{1903}(2^i+5)\equiv7\cdot9\cdot13\cdot(-1)^{19}\equiv-69$.
$(1):$ We have $\displaystyle\prod_{i=1}^{50}(2^{50+i}+5)\equiv\prod_{i=1}^{50}(5-2^i)$, and multiplied with $\displaystyle\prod_{i=1}^{50}(2^i+5)$ gives $\prod_{i=1}^{50}(25-4^i)$.
$(2):$ Same procedure as the previous step.
$(3):$ Using $15625\equiv0\pmod{125}$ and factoring out an odd number of negative signs.
$(4):$ Using $a^b\cdot a^c=a^{b+c}$ and $\displaystyle325=\sum_{i=1}^{25}i$.
$(5):$ Using $4^{25}\equiv-1$, and hence $16^{25}\equiv1$.