Weighing up the positives and the negatives

Let $a$ be the number of $\pm$ symbols that are positive and $b$ be the number of $\pm$ symbols that are $-$ . Since there are are $2018$ $\pm$ symbols, $a + b = 2018$ , and so $b = 2018 - a$ (for integers $a$ and $b$ ).

The equation then becomes $a(x - 1) - b(x - 1) = 2018$ or $(a - b)(x - 1) = 2018$ . Substituting $b = 2018 - a$ gives $(a - (2018 - a))(x - 1) = 2018$ or $(2a - 2018)(x - 1) = 2018$ for integers $a$ and $x$ .

Since $2a - 2018$ is always even for integers $a$ , and $x - 1$ is always an integer for integers $x$ , the number of solutions to the equation would be the number of even factors (positive or negative) of $2018$ . There are only $4$ even factors of $2018$ ( $-2018$ , $-2$ , $2$ , and $2018$ ) and so there are $\boxed{4}$ integer solutions of $x$ .

Let some number of $(x-1)$ 's remain after doing the cancellations. Since there was an even number of them to start with, and we cancel 2 of them at a time, there is an even number of $(x-1)$ 's remaining. Let that number be $2k$ , where $k$ is an integer. Note that $k$ can be negative as well, in which case $-(x-1)$ 's remain. Hence, $2k(x-1)=2018$ , or $k(x-1)=1009$ . Therefore $(x-1)$ is a factor of $1009$ , which is prime. Hence $(x-1)$ can take on $1$ , $-1$ , $1009$ and $-1009$ as its values, so the answer is $\boxed{4}$ .

Each $\pm(x-1)\pm(x-1)$ pair can either equal $0$ or $\pm2(x-1)$ . This means the above equation can be expressed $2y(x-1)=2018$ , where $y$ is the number of positive pairs, minus the number of negative pairs.

$2(x-1)$ must be an even factor of $2018$ , since if it is odd, $x$ will not be an integer, and if it is not a factor of $2018$ , $y$ will not be an integer. This leaves us with the numbers $2$ & $2018$ , and when $y$ is negative, $-2$ & $-2018$ .

This means there are $\boxed{4}$ possible integer values of $x$ : $-1008$ , $0$ , $2$ and $1010$ .

2018=2×2×503

x-1=1/2/1006/503

Let $u = x-1$

Then the above statement is equivalent to saying:

$2nu = 2018$

where $n$ is an integer between $-1009$ and $1009$ .

Therefore, $u$ must be divisible by $2018$ which has only two prime factors $2$ and $1009$ .

Since u can be negative, it can take on only $4$ values, $\pm 2$ and $\pm 1009$ .

Therefore, $x$ can also only take on $\boxed4$ values.

My solution could be wrong, but I would love it someone tells me if it is correcr or wrong and why:

The expression (x-1) is actually just one number if you substitute x so the calculation gives us a pattern of the same number added or subtracted 2017 times to the starting term. 2018 has to be reached in the end.

All the added numbers have to equal 2018 (otherwise we would get 0 or something) so if we only add the number x would have to be 2. If we increase x, we can subtract in the end to cancel out the amount of extra additions but in the end we have to end at 2018.

Therefore I imagined it like climbing stairs were every stair up is +(x-1) and every stair down is -(x-1). We start at 0 and can either take 2018 small stairs only up to get to 2018, or higher steps to go above 2018 and then go back again.

But one step has to go by precisely at 2018 - the height of the stairs therefore has to equal a divisor of 2018. And as 2018 has four divisors: 1, 2, 1009 and 2018 we can find 4 different x to satisfy the equation. Right?