Solving in Integer sequences Each $x_n$ comes from the set $\{2,3,6\}$, 
these statements are true
$x_1 + x_2 + x_3+\cdots+x_n = 633$ 
$\frac{1}{{x_1}^2} + \frac{1}{{x_2}^2} + \frac{1}{{x_3}^2}+\cdots+\frac{1}{{x_n}^2} = \frac{2017}{36}$
What is the value of $n$?
I tried making the number of $2$'s = $a$ , the number of $3$'s = $b$, the number of $6$'s = $c$ , $a+b+c =n$.
I got : 
$2a+3b+6c = 633$
$9a + 4b + c = 2017$
Which I can't solve, I've been guessing and checking and I still did not get it
 A: You are on the right track. After fixing a typo,
$$2a+3b+6c = 633,
\\9a + 4b + c = 2017$$ and by eliminating $a$,
$$19b+52c=1663.$$
The positive solution of this linear equation is $b=41,c=17$, then $a=204$.
A: Your equations are correct, but we can also include the equation $a+b+c=n$.
 
 Solving the system
 $$
 \begin{cases}
 2a+3b+4c=633\\[4pt]
 9a+4b+c=2017\\[4pt]
 a+b+c=n
  \end{cases}
  $$
for $a,b,c,\;$and then clearing denominators, we get the equivalent system
 $$
 \begin{cases}
4a=-7n+2650\\[4pt]
3b=13n-3283\\[4pt]
12c=-19n+5182
  \end{cases}
  $$
In order for $a,b,c\;$to be nonnegative, we must have 


*

*$-7n+2650 \ge 0$, hence $n \le 378$.$\\[4pt]$

*$13n-3283 \ge 0$, hence $n \ge 253$.$\\[4pt]$

*$-19n+5182 \ge 0$, hence $n \le 272$.


Thus, we must have $252\le n\le 272$.

Other than that, we only need to choose $n$ so that


*

*$-7n+2650$ is a multiple of $4$.$\\[4pt]$

*$13n-3283$ is a multiple of $3$.$\\[4pt]$

*$-19n+5182$ is a multiple of $12$.


Solving the associated congruences,


*

*$-7n+2650\equiv 0\;(\text{mod}\;4)$ solves as $n\equiv 2\;(\text{mod}\;4)$.$\\[4pt]$

*$13n-3283\equiv 0\;(\text{mod}\;3)$ solves as $n\equiv 1\;(\text{mod}\;3)$.$\\[4pt]$

*$-19n+5182\equiv 0\;(\text{mod}\;12)$ solves as $n\equiv 10\;(\text{mod}\;12)$.


Noting that $n\equiv 10\;(\text{mod}\;12)$ solves all $3$ congruences, the necessary and sufficient conditions on $n$ are


*

*$253\le n\le 272$$\\[4pt]$

*$n\equiv 10\;(\text{mod}\;12)$


Since $253\equiv 1\;(\text{mod}\;12)$, the least qualifying value of $n$ (in fact, the only qualifying value of $n$) is
$$n=253+9=262$$
