Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$. Which of the following must be a divisor of $a$?

$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$
1. We start by analyzing the given conditions:
   \[
   \gcd(a, b) = 24 = 2^3 \cdot 3
   \]
   \[
   \gcd(b, c) = 36 = 2^2 \cdot 3^2
   \]
   \[
   \gcd(c, d) = 54 = 2 \cdot 3^3
   \]
   \[
   70 < \gcd(d, a) < 100
   \]

2. We express \(a, b, c,\) and \(d\) in terms of their prime factors:
   \[
   a = 2^3 \cdot 3 \cdot a_1 = 24a_1
   \]
   \[
   b = 2^3 \cdot 3^2 \cdot b_1 = 72b_1
   \]
   \[
   c = 2^2 \cdot 3^3 \cdot c_1 = 108c_1
   \]
   \[
   d = 2 \cdot 3^3 \cdot d_1 = 54d_1
   \]

3. We need to find \(\gcd(a, d)\):
   \[
   \gcd(a, d) = \gcd(24a_1, 54d_1)
   \]
   Since \(24 = 2^3 \cdot 3\) and \(54 = 2 \cdot 3^3\), the common factors are \(2\) and \(3\):
   \[
   \gcd(a, d) = 2 \cdot 3 \cdot \gcd(a_1, d_1) = 6 \cdot \gcd(a_1, d_1)
   \]
   Let \(\gcd(a_1, d_1) = n\), then:
   \[
   \gcd(a, d) = 6n
   \]
   Given:
   \[
   70 < 6n < 100
   \]
   Solving for \(n\):
   \[
   \frac{70}{6} < n < \frac{100}{6}
   \]
   \[
   11.67 < n < 16.67
   \]
   Since \(n\) must be an integer:
   \[
   12 \leq n \leq 16
   \]

4. We now test each possible value of \(n\) to see which one fits all conditions:
   - For \(n = 12\):
     \[
     \gcd(a, d) = 6 \cdot 12 = 72 = 2^3 \cdot 3^2
     \]
     This implies \(3 \mid a_1\) and \(4 \mid d_1\). However, this would make \(\gcd(a, b) \geq 72\), which contradicts \(\gcd(a, b) = 24\).

   - For \(n = 13\):
     \[
     \gcd(a, d) = 6 \cdot 13 = 78 = 2 \cdot 3 \cdot 13
     \]
     This implies \(13 \mid a_1\) and \(13 \mid d_1\). This does not contradict any given conditions.

   - For \(n = 14\):
     \[
     \gcd(a, d) = 6 \cdot 14 = 84 = 2^2 \cdot 3 \cdot 7
     \]
     This implies \(7 \mid a_1\) and \(14 \mid d_1\). However, this would make \(\gcd(c, d) \geq 108\), which contradicts \(\gcd(c, d) = 54\).

   - For \(n = 15\):
     \[
     \gcd(a, d) = 6 \cdot 15 = 90 = 2 \cdot 3^2 \cdot 5
     \]
     This implies \(5 \mid a_1\). However, this would make \(\gcd(a, b) \geq 72\), which contradicts \(\gcd(a, b) = 24\).

   - For \(n = 16\):
     \[
     \gcd(a, d) = 6 \cdot 16 = 96 = 2^5 \cdot 3
     \]
     This implies \(16 \mid d_1\). However, this would make \(\gcd(c, d) \geq 108\), which contradicts \(\gcd(c, d) = 54\).

5. The only value of \(n\) that works is \(n = 13\), which implies \(13 \mid a_1\). Therefore, \(13\) must be a divisor of \(a\).

The final answer is \(\boxed{13}\).