Integers a, b, c, d, and e satisfy the following three properties: 
(i) $2 \le a < b <c <d <e <100$ 
(ii)$ \gcd (a,e) = 1 $
(iii) a, b, c, d, e form a geometric sequence. 
What is the value of c?
1. Given that \(a, b, c, d, e\) form a geometric sequence, we can express the terms as:
   \[
   a, ar, ar^2, ar^3, ar^4
   \]
   where \(r\) is the common ratio.

2. From the problem, we know:
   \[
   2 \le a < b < c < d < e < 100
   \]
   and
   \[
   \gcd(a, e) = 1
   \]
   Since \(e = ar^4\), we have:
   \[
   \gcd(a, ar^4) = 1
   \]
   This implies that \(a\) and \(r\) must be coprime.

3. To find \(a\) and \(r\) such that \(a, ar, ar^2, ar^3, ar^4\) are all integers and \(2 \le a < ar < ar^2 < ar^3 < ar^4 < 100\), we need to consider possible values of \(a\) and \(r\).

4. Let's assume \(a = k^4\) for some integer \(k\). This is because if \(r\) is a fraction with a denominator that is \(k\), then \(a\) being \(k^4\) would help in eliminating common factors, ensuring \(\gcd(a, e) = 1\).

5. We need to find \(k\) such that \(a = k^4\) and \(ar^4 < 100\). Let's test \(k = 2\):
   \[
   a = 2^4 = 16
   \]
   If \(a = 16\), then we need to find \(r\) such that:
   \[
   16, 16r, 16r^2, 16r^3, 16r^4
   \]
   are all integers and \(16r^4 < 100\).

6. Let's try \(r = \frac{3}{2}\):
   \[
   16 \left(\frac{3}{2}\right)^4 = 16 \cdot \frac{81}{16} = 81
   \]
   This satisfies \(16 < 24 < 36 < 54 < 81 < 100\).

7. Therefore, the sequence is:
   \[
   16, 24, 36, 54, 81
   \]
   and the value of \(c\) is:
   \[
   c = 36
   \]

The final answer is \(\boxed{36}\)