Hash functions producing uniform outputs
What are the quadratic residues in $\mathbb{Z}_{10}$? What about the cubic residues? What are the orders of $[11]$ and $[12]$ in the group $(\mathbb{Z}_{10}, +)$?

A brief solution follows.

For the sake of clarity, we rename the four functions $a, b, c, d$. Let's analyze their range. Because of the closure properties of the modulo operation, we may do so in the ring $(\mathbb{Z}_{10}, +, \cdot)$. We will prove that $a, c, d$ have a range that is a strict subset of $\mathbb{Z}_{10}$, and can not therefore be uniform, while $b$ is a permutation of $\mathbb{Z}_{10}$. 

Since we are working over a finite ring, we could compute each of the functions on each element of $\mathbb{Z}_{10}$ and prove our claim by exhaustion. Even though that's probably what the exercise wanted you to do, and what you should fall back on if you don't understand what follows, a true gentlemen never does that. Let's work out a more elegant way. 

To begin with, we observe that $a(i) = i^2$ cannot be injective, and therefore, by piegonhole, not even surjective modulo any $m &gt; 2$ as $[1] \neq [-1]$ but $[1]^2 = [1] = [-1]^2$. This also rules out $c$, since $[11] = [1]$ in $\mathbb{Z}_{10}$.

$d$ has to go as well; this stems from the more general fact, which I leave for you to prove, that if $\gcd(k, m) \neq 1$ then $f(i) = ki$ cannot be injective modulo $m$. (Hint: use the Bezout identity).

Now all is left is to prove that $b(i) = i^3$ is a permutation of $\mathbb{Z}_{10}$. First of all, we observe that the ring $\mathbb{Z}_{10}$ is isomorphic to the direct sum of $\mathbb{Z}_5$ and $\mathbb{Z}_2$. Since $b$ is clearly a permutation of $\mathbb{Z}_2$, if we prove that it's also a permutation of $\mathbb{Z}_5$ we are done. But $5$ is prime, so $i^3 \equiv i^{-1}$, which proves our claim.