Emily Thorne is throwing a Memorial Day Party to kick off the Summer in the Hamptons, and she is trying to figure out the seating arrangment for all of her guests. Emily saw that if she seated $4$ guests to a table, there would be $1$ guest left over (how sad); if she seated $5$ to a table, there would be $3$ guests left over; and if she seated $6$ to a table, there would again be $1$ guest left over. If there are at least $100$ but no more than $200$ guests (because she’s rich and her house is $20000$ square feet), what is the greatest possible number of guests?

[i]2015 CCA Math Bonanza Team Round #5[/i]
To solve this problem, we need to find a number \( n \) that satisfies the following conditions:
1. \( n \equiv 1 \pmod{4} \)
2. \( n \equiv 3 \pmod{5} \)
3. \( n \equiv 1 \pmod{6} \)
4. \( 100 \leq n \leq 200 \)

We will use the Chinese Remainder Theorem (CRT) to solve this system of congruences.

1. **Combine the congruences \( n \equiv 1 \pmod{4} \) and \( n \equiv 1 \pmod{6} \):**

   Since both congruences have the same remainder, we can combine them:
   \[
   n \equiv 1 \pmod{\text{lcm}(4, 6)}
   \]
   The least common multiple of 4 and 6 is 12, so:
   \[
   n \equiv 1 \pmod{12}
   \]

2. **Combine the result with \( n \equiv 3 \pmod{5} \):**

   We now have the system:
   \[
   \begin{cases}
   n \equiv 1 \pmod{12} \\
   n \equiv 3 \pmod{5}
   \end{cases}
   \]

   Let \( n = 12k + 1 \) for some integer \( k \). Substitute this into the second congruence:
   \[
   12k + 1 \equiv 3 \pmod{5}
   \]
   Simplify the congruence:
   \[
   12k + 1 \equiv 3 \pmod{5} \implies 2k + 1 \equiv 3 \pmod{5} \implies 2k \equiv 2 \pmod{5} \implies k \equiv 1 \pmod{5}
   \]
   Therefore, \( k = 5m + 1 \) for some integer \( m \). Substitute back into \( n \):
   \[
   n = 12(5m + 1) + 1 = 60m + 13
   \]

3. **Find \( n \) in the range \( 100 \leq n \leq 200 \):**

   We need \( 60m + 13 \) to be within the range:
   \[
   100 \leq 60m + 13 \leq 200
   \]
   Subtract 13 from all parts:
   \[
   87 \leq 60m \leq 187
   \]
   Divide by 60:
   \[
   1.45 \leq m \leq 3.12
   \]
   Since \( m \) must be an integer, the possible values for \( m \) are 2 and 3.

   - For \( m = 2 \):
     \[
     n = 60(2) + 13 = 120 + 13 = 133
     \]
   - For \( m = 3 \):
     \[
     n = 60(3) + 13 = 180 + 13 = 193
     \]

   Both values are within the range, but we need the greatest possible number of guests.

Conclusion:
\[
\boxed{193}
\]