For any real number $$ x $$, let $$ \lfloor x \rfloor $$ be the largest integer less than or equal to $$ x $$. If
$$
\sum_{n=1}^{N} \left\lfloor \frac{1}{5} + \frac{n}{25} \right\rfloor = 25,
$$
then $$ N $$ is:

Question

For any real number $$ x $$, let $$ \lfloor x \rfloor $$ be the largest integer less than or equal to $$ x $$. If  
$$
\sum_{n=1}^{N} \left\lfloor \frac{1}{5} + \frac{n}{25} \right\rfloor = 25,
$$
then $$ N $$ is:

Accepted Answer

To solve the problem:

1. **Expression Simplification**:  
   The given expression is:  
   $$
   \left\lfloor \frac{1}{5} + \frac{n}{25} \right\rfloor = \left\lfloor \frac{n + 5}{25} \right\rfloor.
   $$

2. **Range Analysis**:  
   The value of $$ \left\lfloor \frac{n+5}{25} \right\rfloor $$ changes as $$ n + 5 $$ crosses multiples of 25.  
   - For $$ 1 \leq n \leq 19 $$:  
     $$ n + 5 < 25 $$, so $$ \left\lfloor \frac{n+5}{25} \right\rfloor = 0. $$  
     Count of terms: 19.  
   - For $$ 20 \leq n \leq 44 $$:  
     $$ 25 \leq n + 5 < 50, $$ so $$ \left\lfloor \frac{n+5}{25} \right\rfloor = 1. $$  
     Count of terms: 25.

3. **Summation Check**:  
   The sum becomes:  
   - From $$ n = 1 $$ to $$ 19: 19(0) = 0. $$  
   - From $$ n = 20 $$ to $$ 44: 25(1) = 25. $$

4. **Conclusion**:  
   At $$ N = 44, \sum_{n=1}^{N} = 0 + 25 = 25. $$

Thus, the value of $$ N $$ is **44**.