Vtyjkyuk

Each $x_n$ comes from the set $2,3,6$,
these statements are true

$x_1 + x_2 + x_3+cdots+x_n = 633$

$frac1x_1^2 + frac1x_2^2 + frac1x_3^2+cdots+frac1x_n^2 = frac201736$

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

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 $a=204$.

Your equations are correct, but we can also include the equation $a+b+c=n$.

Solving the system
$$
begincases
2a+3b+4c=633\[4pt]
9a+4b+c=2017\[4pt]
a+b+c=n
endcases
$$
for $a,b,c,;$and then clearing denominators, we get the equivalent system
$$
begincases
4a=-7n+2650\[4pt]
3b=13n-3283\[4pt]
12c=-19n+5182
endcases
$$
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 $252le nle 272$.

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,

• $-7n+2650equiv 0;(textmod;4)$ solves as $nequiv 2;(textmod;4)$.$\[4pt]$
  


• $13n-3283equiv 0;(textmod;3)$ solves as $nequiv 1;(textmod;3)$.$\[4pt]$
  


• $-19n+5182equiv 0;(textmod;12)$ solves as $nequiv 10;(textmod;12)$.
  

Noting that $nequiv 10;(textmod;12)$ solves all $3$ congruences, the necessary and sufficient conditions on $n$ are

• $253le nle 272$$\[4pt]$
  


• $nequiv 10;(textmod;12)$
  

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