Let $d$ be the greatest common divisor of $2^{30^{10}}-2$ and $2^{30^{45}}-2$.  Find the remainder when $d$ is divided by $2013$.
1. We start with the given problem: finding the greatest common divisor (gcd) of \(2^{30^{10}} - 2\) and \(2^{30^{45}} - 2\). We denote this gcd by \(d\).

2. Using the property of gcd for numbers of the form \(a^m - a\) and \(a^n - a\), we have:
   \[
   \gcd(a^m - a, a^n - a) = a^{\gcd(m, n)} - a
   \]
   Here, \(a = 2\), \(m = 30^{10}\), and \(n = 30^{45}\). Therefore:
   \[
   d = \gcd(2^{30^{10}} - 2, 2^{30^{45}} - 2) = 2^{\gcd(30^{10}, 30^{45})} - 2
   \]

3. Next, we need to find \(\gcd(30^{10}, 30^{45})\). Since \(30 = 2 \times 3 \times 5\), we can use the property of gcd for powers:
   \[
   \gcd(30^{10}, 30^{45}) = 30^{\gcd(10, 45)}
   \]
   The gcd of 10 and 45 is 5, because:
   \[
   \gcd(10, 45) = 5
   \]
   Therefore:
   \[
   \gcd(30^{10}, 30^{45}) = 30^5
   \]

4. Substituting back, we get:
   \[
   d = 2^{30^5} - 2
   \]

5. Now, we need to find the remainder when \(d\) is divided by 2013. First, we compute \(\phi(2013)\), where \(\phi\) is the Euler's totient function. Since \(2013 = 3 \times 11 \times 61\), we have:
   \[
   \phi(2013) = \phi(3) \times \phi(11) \times \phi(61) = (3-1)(11-1)(61-1) = 2 \times 10 \times 60 = 1200
   \]

6. Since \(\phi(2013) = 1200\), we know that \(2^{1200} \equiv 1 \pmod{2013}\) by Euler's theorem. We need to reduce the exponent \(30^5\) modulo 1200:
   \[
   30^5 \mod 1200
   \]
   Calculating \(30^5\):
   \[
   30^5 = 24300000
   \]
   Now, reducing modulo 1200:
   \[
   24300000 \mod 1200 = 0
   \]
   because \(24300000\) is a multiple of 1200.

7. Therefore:
   \[
   2^{30^5} \equiv 2^0 \equiv 1 \pmod{2013}
   \]

8. Finally:
   \[
   d = 2^{30^5} - 2 \equiv 1 - 2 \equiv -1 \equiv 2012 \pmod{2013}
   \]

The final answer is \(\boxed{2012}\).