If the integers $m,n,k$ hold the equation $221m+247n+323k=2001$, what is the smallest possible value of $k$ greater than $100$?

$ 
\textbf{(A)}\ 124
\qquad\textbf{(B)}\ 111
\qquad\textbf{(C)}\ 107
\qquad\textbf{(D)}\ 101
\qquad\textbf{(E)}\ \text{None of the preceding}
$
1. We start with the given equation:
   \[
   221m + 247n + 323k = 2001
   \]

2. To find the smallest possible value of \( k \) greater than 100, we will consider the equation modulo 13. First, we reduce each coefficient modulo 13:
   \[
   221 \equiv 0 \pmod{13} \quad \text{(since } 221 = 17 \times 13)
   \]
   \[
   247 \equiv 0 \pmod{13} \quad \text{(since } 247 = 19 \times 13)
   \]
   \[
   323 \equiv -2 \pmod{13} \quad \text{(since } 323 = 25 \times 13 - 2)
   \]
   \[
   2001 \equiv 12 \pmod{13} \quad \text{(since } 2001 = 154 \times 13 + 12)
   \]

3. Substituting these congruences into the original equation, we get:
   \[
   0m + 0n - 2k \equiv 12 \pmod{13}
   \]
   Simplifying, we have:
   \[
   -2k \equiv 12 \pmod{13}
   \]

4. To solve for \( k \), we need to isolate \( k \). First, we multiply both sides by the modular inverse of \(-2\) modulo 13. The modular inverse of \(-2\) modulo 13 is 7 because:
   \[
   -2 \times 7 \equiv -14 \equiv -1 \equiv 12 \pmod{13}
   \]
   Therefore:
   \[
   k \equiv 12 \times 7 \pmod{13}
   \]
   \[
   k \equiv 84 \pmod{13}
   \]
   \[
   k \equiv 7 \pmod{13}
   \]

5. We need the smallest \( k \) greater than 100 that satisfies \( k \equiv 7 \pmod{13} \). We can write \( k \) in the form:
   \[
   k = 13n + 7
   \]
   where \( n \) is an integer. To find the smallest \( k \) greater than 100:
   \[
   13n + 7 > 100
   \]
   \[
   13n > 93
   \]
   \[
   n > \frac{93}{13}
   \]
   \[
   n > 7.1538
   \]
   The smallest integer \( n \) greater than 7.1538 is 8. Therefore:
   \[
   k = 13 \times 8 + 7 = 104 + 7 = 111
   \]

The final answer is \(\boxed{111}\).