Consider the base 27 number 
\[
  n = ABCDEFGHIJKLMNOPQRSTUVWXYZ ,
\]	
where each letter has the value of its position in the alphabet.  What remainder do you get when you divide $n$ by 100?  (The remainder is an integer between 0 and 99, inclusive.)

1. **Express the number in base 27:**
   \[
   n = 1 \cdot 27^{25} + 2 \cdot 27^{24} + 3 \cdot 27^{23} + \ldots + 25 \cdot 27 + 26
   \]

2. **Rewrite the sum:**
   \[
   n = (27^{25} + 27^{24} + \ldots + 27 + 1) + (27^{24} + 27^{23} + \ldots + 27 + 1) + \ldots + (27 + 1) + 1
   \]

3. **Use the geometric series formula:**
   Each term in the sum can be expressed using the geometric series formula:
   \[
   \sum_{k=0}^{m} 27^k = \frac{27^{m+1} - 1}{26}
   \]
   Applying this to each term, we get:
   \[
   n = \frac{27^{26} - 1}{26} + \frac{27^{25} - 1}{26} + \frac{27^{24} - 1}{26} + \ldots + \frac{27^2 - 1}{26} + \frac{27 - 1}{26}
   \]

4. **Combine the terms:**
   \[
   26n = 27^{26} + 27^{25} + 27^{24} + \ldots + 27^2 + 27 - 26
   \]

5. **Use the geometric series formula again:**
   \[
   26n = \frac{27^{27} - 1}{26} - 27
   \]
   Simplifying, we get:
   \[
   26n = \frac{27^{27} - 703}{26}
   \]
   Multiplying both sides by 26:
   \[
   676n = 27^{27} - 703
   \]

6. **Simplify using modular arithmetic:**
   Since \( \lambda(100) = 20 \) (Carmichael function), we have:
   \[
   3^{81} \equiv 3 \mod 100
   \]
   Therefore:
   \[
   27^{27} \equiv 3^{81} \equiv 3 \mod 100
   \]
   Thus:
   \[
   676n \equiv 3 - 703 \equiv 3 - 703 \equiv -700 \equiv 0 \mod 100
   \]

7. **Determine possible values of \( n \mod 100 \):**
   Since \( 676n \equiv 0 \mod 100 \) and \( \gcd(676, 100) = 4 \), we have:
   \[
   676n \equiv 0 \mod 100 \implies n \equiv 0, 25, 50, 75 \mod 100
   \]

8. **Use modulo 4 to narrow down the options:**
   \[
   n \equiv (2+4+6+\ldots+26) + 3(1+3+5+\ldots+25) \mod 4
   \]
   The sum of the first series is:
   \[
   2(1+2+3+\ldots+13) = 2 \cdot \frac{13 \cdot 14}{2} = 13 \cdot 14 \equiv 1 \mod 4
   \]
   The sum of the second series is:
   \[
   3(1+3+5+\ldots+25) = 3 \cdot 13^2 = 3 \cdot 169 \equiv 3 \cdot 1 = 3 \mod 4
   \]
   Therefore:
   \[
   n \equiv 1 + 3 = 4 \equiv 0 \mod 4
   \]
   This eliminates 25, 50, and 75, leaving:
   \[
   n \equiv 0 \mod 100
   \]

The final answer is \( \boxed{0} \)