Modular Inverse Problem What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{1050}$ is defined?
How do you do modular inverses?

Given a number $n$, if a number $m$ exists such that $n\cdot m \equiv 1\pmod{1050}$ then $m$ is said to be a modular inverse for $n$ and could have been written as $n^{-1}$.
Moving to a bit of group theory, in terms of addition, the equivalence classes of the modulo 1050 operation form a group, commonly written as either $\mathbb{Z}/1050\mathbb{Z}$ or as $\mathbb{Z}_{1050}$, where the group operation is addition.  When taken with multiplication as well, this forms a ring.
The group of units of such a ring is often denoted with a multiplication symbol at the top right, such as $\mathbb{Z}_{1050}^\times$, and is defined as those elements in $\mathbb{Z}_{1050}$ that have multiplicative inverses which in particular implies that they form a multiplicative group.
$\mathbb{Z}_{k}^\times$ in general will consist of all elements $1\leq n\leq k$ such that $n$ is coprime to $k$ (i.e. $\gcd(n,k)=1$).  It is important to note that $\mathbb{Z}_p^\times = \{1,2,\dots,p-1\}$ for $p$ prime since every element has a multiplicative inverse except for zero.  This is the prototypical example of a finite field. 
For example, $\mathbb{Z}_8^\times = \{1,3,5,7\}$ which is isomorphic to the Klein-4 group (since you have $3\times 3 \equiv 1\pmod{8}$, $5\times 5\equiv 1\pmod{8}$ and $7\times 7\equiv 1\pmod{8}$)
The result then is that the smallest integer $n>1$ such that $n^{-1}$ is defined is equivalent to asking what is the smallest $n>1$ such that $\gcd(n,1050)=1$.
Actually calculating what the modular inverse of an element is can be done using the euclidean division algorithm.  $\gcd(n,k)=1$ implies that there exist $a$ and $b$ such that $an+bk = 1$.  Keeping track of your steps taken in the division algorithm process will tell you what the values of $a$ and $b$ actually are.  In trying to find the modular inverse of $n$ in $\mathbb{Z}_k^\times$, taking the equation $1=an+bk$ and modding out by $k$, you are left with $an\equiv 1\pmod{k}$ and so you would have $a$ as your modular inverse.

As you figured out in the comments below, the answer will indeed by $n=11$ (since each number $2,\dots,10$ share at least one factor with $1050$).  Here, we calculate what $11^{-1}$ actually is.  We run the Euclidean division algorithm while keeping track of the steps used.
$\begin{array}{l|l|l}
1050&1&0\\
11&0&1\\
5&1&-95\\
1&-2&191\end{array}$
Each line can be read as an equation where the first column is what it equals, the second column is the coefficient of $1050$, and the third column is the coefficient of $11$.  The third line for example reads $5=1\cdot 1050 + (-95)\cdot 11$.  The final line reads $1 = -2\cdot 1050 + 191\cdot 11$.
Since $1$ is equal to $-2\cdot 1050 + 191\cdot 11$, they are certainly also equivalent when considering them modulo $1050$.  We have then $1\equiv -2\cdot 1050+191\cdot 11 \equiv 191\cdot 11\pmod{1050}$ showing that $191=11^{-1}$.
