A number theory problem by Vu Vincent

Each term can be written as $t_n=4n-1$ where $1 \leq n \leq 503$

When $n=2k$ where $n=2,4,6,8,....,502$ then $t_n=8k-1 \equiv -1 \pmod{8}$

When $n=2k-1$ where $k=1,3,5,.....,503$ then $t_n=4(2k-1)-1=8k-5 \equiv 3 \pmod{8}$

$A \equiv (-1)^{251} \times 3^{252} \equiv -1 \times 9^{126} \equiv -1 \equiv 7 \pmod{8}$

Next part we have to find the remainder when $A$ is divided by $125$

Clearly, $A$ has a term $375=4.94-1$ which is divisible by $125$ .So, $A$ is divisible by $125$

After summarizing,

$A \equiv 7 \pmod{8} \\ A \equiv 0 \pmod{125}$

After getting some trials it can be found that only $375$ satisfies the above two conditions.

Therefore, $A \equiv 375 \pmod{1000}$