Algebraic Locks (Set $1$ , Problem $1$ , Version $2$ )

Since $y_1 + 8 = y_3$ , $y_1 = 1, y_3 = 9$

Since $y_4 + 2(y_2) = 9$ , we need to check all the permutations of the sums of $9$ :

$1 + 8 = 9$

$2 + 7 = 9$

$3 + 6 = 9$

$4 + 5 = 9$

$5 + 4 = 9$

$6 + 3 = 9$

$7 + 2 = 9$

$8 + 1 = 9$

Since $y_4, y_2$ are different digits, we remove the ones that results in the violation of the different digits rule:

$1 + 8 = 9$

$2 + 7 = 9$

$4 + 5 = 9$

$5 + 4 = 9$

$7 + 2 = 9$

$8 + 1 = 9$

Now, we remove the ones that cannot support the $2(y_2)$ part of the second clue:

$1 + 8 = 9$

$4 + 5 = 9$

$5 + 4 = 9$

$7 + 2 = 9$

Now, we remove the ones that reverses the $y$ -values in the second clue:

$1 + 8 = 9$

$5 + 4 = 9$

$7 + 2 = 9$

Now, remove the ones that violate the different digits rule again:

$5 + 4 = 9$

Therefore $y_4 = 5, y_2 = 2$

Now, plug the $y$ -values in:

$x + 1 + x - 2 + x + 9 + x - 5 = 4x + 10 - 7 = 4x + 3 = x$

Simplify:

$3x + 3 = 0$

$3x = -3$

$x = -1$ - remove the $-$ sign - $x = 1$

Plug $x = 1$ :

$1 + 1 = 2$

$1 - 2 = -1$ - remove the $-$ sign - $1 - 2 = 1$

$1 + 9 = 10 = 0$

$1 - 5 = -4$ - remove the $-$ sign - $1 - 5 = 4$

Code = $\fbox {2104}$