Divisibility remainders for $9, 25, 49,$ and $105$.

Question

An integer $n$ satisfies $n$ is congruent to $5 \pmod 9$, $n$ is congruent to $12 \pmod{25}$, and $n$ is congruent to $44 \pmod{49}$. What is the remainder when $n$ is divided by $105$? Find all possibilities.

Answer 1

Assuming you meant divided by $105$ as in the title, the equivalences

$$n\equiv5\pmod9,n\equiv12\pmod{25},n\equiv44\pmod{49}$$

can be reduced to

$$n\equiv 2\pmod3,n\equiv2\pmod5,n\equiv2\pmod7$$.

From here it's easy to see that $n\equiv2\pmod{3\times5\times7}$.

Answer 2

Using Chinese remainder theorem, we get $$n\equiv 6512\pmod{105^2}$$

Since $6512=62\cdot105+2$ then $n\equiv 2\pmod {105}$.