Find the smallest number such that when you divide with $4,6,8,10,12$ the remainder is $2,4,6,8,10$.
This is the solution:
Let $n \in \mathbb N$ such that $n = 4q_1 + 2$, for some $q_1$, $n = 6q_2 + 4$, for some $q_2$, $n = 8q_3 + 6$, for some $q_3$, $n = 10q_4 + 8$, for some $q_4$, $n = 12q_5 + 10$, for some $q_5$.
Then $4 \mid n + 2$, $6 \mid n + 2$, $8 \mid n + 2$, $10 \mid n + 2$, $12 \mid n + 2$, so $[4, 6, 8, 10, 12] \mid n + 2$ so $120 \mid n + 2$ as $[4, 6, 8, 10, 12] = 120$ so the smallest number is $118$.
I do not understand how he get that $4 \mid n + 2$, $6 \mid n + 2$, $8 \mid n + 2$, $10 \mid n + 2$, $12 \mid n + 2$, can you explain it to me?
$$n+2=4q_1+2+2=4(q_1+1)\\n+2=6q_2+4+2=6(q_2+1)\\n+2=8q_3+6+2=8(q_3+1)\\\cdots$$