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.
Divisibility remainders for $9, 25, 49,$ and $105$.
1
$\begingroup$
modular-arithmetic
divisibility
-
0Have you ... tried using the Chinese Remainder Theorem? – 2017-01-05
-
4The title says $105$, but it is not in the text of your question. Did you mean "...when $n$ id divided by $105$?" – 2017-01-05
-
0@ajotatxe Good Catch. I changed it – 2017-01-05
2 Answers
3
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 $n\equiv2\pmod{3\times5\times7}$.
1
Using Chinese remainder theorem, we get $$n\equiv 6512\pmod{105^2}$$
Since $6512=62\cdot105+2$ then $n\equiv 2\pmod {105}$.