Sum of Greatest Integer Functions

Let $t_n = \left[\frac{1}{5} + \frac{n}{25}\right] = \left[\frac{5 + n}{25}\right]$.

For $n = 1, 2, \dots, 19$: $\frac{5 + n}{25} \in \left[\frac{6}{25}, \frac{24}{25}\right] \implies t_n = 0$

For $n = 20, 21, \dots, 44$: $\frac{5 + n}{25} \in \left[\frac{25}{25}, \frac{49}{25}\right] = [1, 1.96] \implies t_n = 1$

Sum for $n = 1$ to $N$:

• For $1 \le n \le 19$, sum is $0$.
• For $n = 20$ onwards, each term contributes $1$. We need the sum to equal $25$. Number of terms with $t_n = 1$ needed is $25$. Since $t_n = 1$ starts at $n = 20$, the 25th such term is $n = 20 + 25 - 1 = 44$.

Thus, $N = 44$.