How many of the numbers $2,6,12,20,\ldots,14520$ are divisible by $120$?

$\text{(A) }2\qquad\text{(B) }8\qquad\text{(C) }12\qquad\text{(D) }24\qquad\text{(E) }32$
1. **Identify the sequence and its general term:**
   The given sequence is \(2, 6, 12, 20, \ldots, 14520\). We notice that each term can be written as \(n(n+1)\) where \(n\) is a positive integer. For example:
   - \(2 = 1 \cdot 2\)
   - \(6 = 2 \cdot 3\)
   - \(12 = 3 \cdot 4\)
   - \(20 = 4 \cdot 5\)
   - \(\ldots\)
   - \(14520 = 120 \cdot 121\)

2. **Determine the condition for divisibility by 120:**
   We need to find how many terms \(n(n+1)\) are divisible by \(120\). Since \(120 = 2^3 \cdot 3 \cdot 5\), we need:
   \[
   n(n+1) \equiv 0 \pmod{8}, \quad n(n+1) \equiv 0 \pmod{3}, \quad n(n+1) \equiv 0 \pmod{5}
   \]

3. **Analyze each modulus condition:**
   - For \(n(n+1) \equiv 0 \pmod{8}\):
     Since \(n\) and \(n+1\) are consecutive integers, one of them must be even. Therefore, \(n(n+1)\) is always even. To be divisible by \(8\), one of them must be divisible by \(8\) or both must be divisible by \(4\). This gives:
     \[
     n \equiv 0, 7 \pmod{8}
     \]
   - For \(n(n+1) \equiv 0 \pmod{3}\):
     One of \(n\) or \(n+1\) must be divisible by \(3\). This gives:
     \[
     n \equiv 0, 2 \pmod{3}
     \]
   - For \(n(n+1) \equiv 0 \pmod{5}\):
     One of \(n\) or \(n+1\) must be divisible by \(5\). This gives:
     \[
     n \equiv 0, 4 \pmod{5}
     \]

4. **Combine the conditions using the Chinese Remainder Theorem:**
   We need to find \(n\) that satisfies all three conditions simultaneously. We solve the system of congruences:
   \[
   n \equiv 0, 7 \pmod{8}
   \]
   \[
   n \equiv 0, 2 \pmod{3}
   \]
   \[
   n \equiv 0, 4 \pmod{5}
   \]

5. **Count the solutions:**
   We need to find the number of \(n\) values from \(1\) to \(120\) that satisfy these conditions. By the Chinese Remainder Theorem, the number of solutions modulo \(120\) (since \(120 = 8 \cdot 3 \cdot 5\)) is the product of the number of solutions for each modulus condition:
   - For \(n \equiv 0, 7 \pmod{8}\), there are 2 solutions.
   - For \(n \equiv 0, 2 \pmod{3}\), there are 2 solutions.
   - For \(n \equiv 0, 4 \pmod{5}\), there are 2 solutions.

   Therefore, the total number of solutions is:
   \[
   2 \times 2 \times 2 = 8
   \]

The final answer is \(\boxed{8}\).