Assume an encrypted message is sent through use of exponential cipher. ...such that modulus $p = 2741 \text{ (p is prime)}$ and $e = 11 \text{ (e = exponent)}$
Message: $1315\quad 0611 \quad 0427 \quad 0091 \quad 0520 \quad 0733$
I am required to determine the decryption exponent and determine what it says.
This is my work so far on Mathematica. I believe my solution is wrong, as
the exponent I arrived at is negative. Please advise. Thanks in advance.

'In' =input 
'Out' =output
In $ \text{mssg} = { 1315, 0611, 0427, 0091, 0520, 0733}$
Out ${1315, 611, 427, 91, 520, 733}$
In $p = 2741$
In $e = 11$
In $\text {code[x_]} := \mod[x^{e}, p]$
In $\text { cipher  = code[mssg] }$
Out $\, {2622, 2659, 1544, 951, 2718, 859}$
In $\text{ ExtendedGCD }[e, p - 1]$
Out $\,(1, (-249, 1))$

Why do my exponent here shows as $-249 ?$


Unfortunately, The Mathematica online help page about ExtendedGCD gives limited information about this;
\begin{align}
in[1]:&\; \{g, \{a, b\}\} = \operatorname{ExtendedGCD}[2, 3]\\
out[1]:&\; \{1,\{-1,1\}\} \quad \text{ //next, test the result}\\ 
in[2]: &\;2 a + 3 b == g\\
out[2]:&\; \texttt{True}\\
\end{align} 
The extra output sub-list is for the Bezout's Identity.:

Let $a$ and $b$ be integers with greatest common divisor $d$. Then, there exist integers $x$ and $y$ such that $ax + by = d$

For your problem;
\begin{align}
 \gcd(e,p-1) &= u \cdot e + v \cdot (p-1), \text{ for some } u,v\in\mathbb{Z}\\
1  &=  -249 \cdot 11 + 1 \cdot 2740\\
1  &=  -2739 +  2740
\end{align}
Actually, you want the inverse of $e$ $\bmod{p-1}$ and Bezout's Identity that is very helpful to find the inverses.
If you take $\bmod 2740$ on both side - that is one of the ways to calculate the inverse -;
$$ 1  =  -249 \cdot 11 + 1 \cdot 2740$$
$$ 1  \equiv  -249 \cdot 11 \pmod{2740}$$
you fill find the inverse of $11 \pmod{2740}$, that is $-249 \equiv 2491 \bmod 2740$ 
