1. The natural number $n$ when divided by 2 has a remainder $a$, when divided by 4 has a remainder $b$, and when divided by 6 has a remainder $c$. If $a+b+c=9$, determine the remainder when the number $n$ is divided by 12.
Solution. The number $n$ can be written in the form $n=2p+a, 0 \leq a \leq 1$, $n=4q+b, \quad 0 \leq b \leq 3$ and $n=6r+c, \quad 0 \leq a \leq 1$. From this, it follows that $0 \leq a+b+c \leq 9$. Since $a+b+c=9$ it follows that $a=1, b=3$ and $c=5$. Then $n+1=2p+2=4q+4=6r+6$. Thus $n+1$ is divisible by 2, 4, and 6, and therefore is also divisible by 12. Thus $n+1=12k$, or $n=12k-1$. It follows that the remainder when the number $n$ is divided by 12 is 11.