What is the number of elements in $\{1,2,...,2018\}$ such that it can be represented by $ x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \:\:? $ What is the number of elements in $\{1,2,...,2018\}$ such that it can be represented by
$$ x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \:\:? $$
$x \in \mathbb{Z}$

Attempt :
First, $x=1$ then the sum will be $7$, which is an element of the set in concern. Of course $x=0$ then the sum will be $1$, also in the set.
For $x=-1$, the sum will also be $1$. 
So far we have 2 elements of the set.
For $|x|>1$, we can view the summation as a finite geometric series :
$$ S = 1 + x + .... + x^{6} $$
$$ Sx = x + x^{2} + .... + x^{7} $$
subtract both of the above then we can have
$$ S(x) = \frac{1 - x^{7}}{1-x}, \:\:\: |x| > 1 $$
The key is to find $x*$ so that $S(x*)$ is the maximum just below $2018$.
I can do this by checking manually from $x=\pm 2, \pm 3, \pm 4 $. Without a calculator, this is quite tedious. Found out that $x = \pm 4$ do not satisfy the condition. 

So the answer is $6$ numbers, is this sufficient? (I was thinking that we may also have to check the monotonicity of $S(x)$)
What techniques are better than this one? Thanks.
 A: Working from your expression $S=\frac {x^7-1}{x-1}$ or from the original sum it is natural to start by taking the sixth root of $2018$.  Sixth roots are hard, so taking a square root and then a cube root is the way to go.  Maybe you know that $45^2=2025$ which makes it clear that $\pm 3$ will work because $3^3=27$ is so much less than $45$ and leaves you in doubt about $\pm 4$.  We have $4^7=2^{14}$ and maybe you know that $2^{14}=16384$.  Then it is clear that $\pm 4$ leads to values too large and the result is that $7$ values of $x$ give a number in range. Since two of them give the same number the answer to the question is $6$.  If you are going to do things like this without a calculator it helps to have a range of facts available.
A: Your approach to express  $$S(x) = \frac{1 - x^{7}}{1-x}, \:\:\: |x| > 1 $$is a good one.
Since $$4^6 = 4096$$ is obviously too large, you stay with $$x\in \{0, \pm 1,\pm 2, \pm 3\}$$ to get all the answers. 
I found $$\{1,7,43,127,547,1093\}$$
A: For  brevity let $x^6 +x^5+ x^4 +x^3+ x^2+x+1=f(x).$... For $x\in \Bbb Z$ we have:
$$(Ia).\quad |x|\geq 4\implies f(x)=(x^6+x^4+x^2)(1+1/x)+1\geq$$ $$\geq (x^6+x^4+x^2)(1-1/|x|)+1>$$ $$>  x^6(1-1/|x|)\geq$$ $$\geq 4^6(1-1/2)=64^2/2=2048>2018.$$
$$(Ib).\quad |x|\leq 3\implies |f(x)|\leq |x|^6+|x|^5+|x|^4|+|x|^3+|x|^2+|x|+1\leq$$ $$\leq f(3)=(3^7-1)/2=(3^8/3-1)/2=$$ $$=(81^2/3-1)/2=1093<2018.$$
$(II). $ Obviously $x\geq 0\implies f(x)\geq 1.$  And $x<0\implies 1+1/x\geq 0\implies f(x)=(x^6+x^4+x^2)(1+1/x)+1\geq 1.$
$(III). $ So the solution set is $\{0,\pm 1,\pm 2,\pm 3\}.$
A: 6
The question looks hard at first, but it really isn't! First look at the graph,
The graph
The above graph was of $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$, As you can judge from the scale, the y value increases very rapidly, and quickly reaches  $2018$, We can easily put values and see for ourselves the rapidity,
$$ f(0)=1$$
$$ f(1)=7$$
$$ f(2)=127$$
$$ f(3)=1093$$
$f(4)=5461$, which is of course, $>2018$, and hence all the values henceforth will be greater than this, Now, lets tackle negative ones,
$$f(-1)=1$$
$$f(-2)=43$$
$$f(-3)=547$$
$f(-4)=3277$, which is again $>2018$, and from the graph we can tell that, It will keep on increasing further.
Thus the set $S$ containing the elements of $\{1,2,3\ldots2018\}$, which satisfy $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$, where, $x\in \mathbb{Z}$ are,
$$\boxed{\,\,\{1,7,43,127,547,1093\}\,\,}$$
