33. It is given that $a, b$ and $c$ are three positive integers such that
$$
a^{2}+b^{2}+c^{2}=2011 .
$$

Let the highest common factor (HCF) and the least common multiple (LCM) of the three numbers $a, b, c$ be denoted by $x$ and $y$ respectively. Suppose that $x+y=388$. Find the value of $a+b+c$. (Remark: The highest common factor is also known as the greatest common divisor.)
33. Answer: 61
Without loss of generality, we may assume that $a \geq b \geq c$. Let the HCF (or GCD) of $a, b$ and $c$ be $d$. Then $a=d a_{1}, b=d b_{1}$ and $c=d c_{1}$. Let the LCM of $a_{1}, b_{1}$ and $c_{1}$ be $m$. Thus, $a_{1}^{2}+b_{1}^{2}+c_{1}^{2}=\frac{2011}{d}$ and $d+m d=388$ or $1+m=\frac{388}{d}$. So, $d^{2} \mid 2011$ and $d \mid 388$. Note that 2011 is a prime. Thus we must have $d=1$, and it follows that $a=a_{1}, b=b_{1}, c=c_{1}$, and thus $a^{2}+b^{2}+c^{2}=2011$. In particular, $a^{2}+b^{2}+c^{2}<2025=45^{2}$, so that one has $a, b, c<45$. Furthermore we have $m=387=3^{2} \times 43$. Thus $a, b$ and $c$ can only be $1,3,9$ or 43 , since they must be less than 45 . Then it is easy to check that $43^{2}+9^{2}+9^{2}=2011$, and $a=43$, $b=c=9$ is the only combination which satisfies the given conditions. Thus we have $a+b+c=43+18=61$.