Prove that a number is not an exact cube by using modular arithmetic mod 9 My textbook asks me to prove that the number $$1\underbrace{000\ldots0}_{2012}5\underbrace{000\ldots0}_{2012}1$$ 
is not an exact cube of any natural number. And in the answers section it states that one shall use the properties of modulo 9, however I can't understand, why the number 9 is chosen for the task. Why don't we relate modulo 7? Or 11? I.e. why is the number 9 so special that we can use its residues' properties to prove whether a number is a cube?
 A: There is no assertion that $9$ can be used to prove whether any number is a cube. 
There is only an assertion that $9$ can be used to prove that this number is not a cube, which is easy enough to check, because this number is congruent to $7$ modulo $9$, and $7$ is not a cube modulo $9$.
All this problem is saying is that working in arithmetic modulo $9$ is sufficient to prove that this number is not a cube. You can of course explore interesting mathematical questions such as: what other bases $n$ can be used in which to prove that this number is not a cube modulo $n$?
A: Computing modulo 9 has two relevant properties:


*

*In base ten, it is very easy to reduce a large number modulo 9, because the result is just the iterated digit sum.
So the large number in your exercise has residue $1+5+1=7$ modulo $9$.

*The only cubes modulo $9$ are $0$, $1$ and $8$, which you can check simply by taking the cube of each integer from $0$ to $8$. (This can made somewhat faster by noting that the cubes of $5,6,7,8$ are simply the negatives of the cubes of $4,3,2,1$).
Since $7$ is not in $\{0,1,8\}$, the large number is not a perfect cube.

Actually, computing modulo $7$ would work for this case too: The only cubes modulo $7$ are $\{0,1,6\}$, and your number is congruent to $4 \pmod 7$. But finding that $4$ takes quite a bit more effort than just summing digits.
Computing modulo $11$ would not teach us anything, because every residue class modulo $11$ is a possible cube. (This again can be seen by explicit computation, or with a bit of abstract algebra knowledge: $11$ is prime, so $\mathbb Z_{11}$ is a finite field, so its multiplicative group is cyclic of order $10$, which is coprime to $3$. So raising to the third power in $\mathbb Z_{11}$ is a bijection).

As Lee Mosher points out, it is not always the case that modulo $9$ works best. If we want to find out whether
$$ \underbrace{777\ldots777}_{2013}4 $$
is a cube, a modulo-9 test will be inconclusive (the number is congruent to $1$ modulo 9), but a modulo-7 test immediately tells us it can't be a cube.
