 Let $$N=2^5+2^{5^2}+2^{5^3}+\cdots{}+2^{5^{2015}}.$$Wrotten in the usual decimal form, find the last two digits of the number $N$.

To find the last two digits of \( N = 2^5 + 2^{5^2} + 2^{5^3} + \cdots + 2^{5^{2015}} \), we need to compute \( N \mod 100 \).

1. **Compute \( N \mod 4 \):**
   \[
   2^5 \equiv 0 \pmod{4}, \quad 2^{5^2} \equiv 0 \pmod{4}, \quad \ldots, \quad 2^{5^{2015}} \equiv 0 \pmod{4}
   \]
   Since each term in the sum is divisible by 4, we have:
   \[
   N \equiv 0 \pmod{4}
   \]

2. **Compute \( N \mod 25 \):**
   By Euler's Totient Theorem, since \(\phi(25) = 20\), we have:
   \[
   2^{20} \equiv 1 \pmod{25}
   \]
   We need to determine the exponent \( 5^k \mod 20 \):
   \[
   5^1 \equiv 5 \pmod{20}, \quad 5^2 \equiv 25 \equiv 5 \pmod{20}, \quad \ldots, \quad 5^k \equiv 5 \pmod{20} \quad \text{for all } k \geq 1
   \]
   Therefore:
   \[
   2^{5^k} \equiv 2^5 \equiv 32 \equiv 7 \pmod{25}
   \]
   Since there are 2015 terms, we have:
   \[
   N \equiv 2015 \cdot 7 \pmod{25}
   \]
   Simplify \( 2015 \mod 25 \):
   \[
   2015 \div 25 = 80 \quad \text{remainder } 15 \quad \Rightarrow \quad 2015 \equiv 15 \pmod{25}
   \]
   Thus:
   \[
   N \equiv 15 \cdot 7 \equiv 105 \equiv 5 \pmod{25}
   \]

3. **Combine results using the Chinese Remainder Theorem:**
   We have:
   \[
   N \equiv 0 \pmod{4} \quad \text{and} \quad N \equiv 5 \pmod{25}
   \]
   Let \( N = 25k + 5 \). Then:
   \[
   25k + 5 \equiv 0 \pmod{4} \quad \Rightarrow \quad 25k + 5 \equiv k + 1 \equiv 0 \pmod{4} \quad \Rightarrow \quad k \equiv 3 \pmod{4}
   \]
   Let \( k = 4m + 3 \). Then:
   \[
   N = 25(4m + 3) + 5 = 100m + 75 + 5 = 100m + 80
   \]
   Therefore:
   \[
   N \equiv 80 \pmod{100}
   \]

The final answer is \(\boxed{80}\)