Solving in Integer sequences

Question

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

Answer 1

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 $a=204$.

Answer 2

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 $$n=253+9=262$$