Discussion

Question:

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

Explanation:

$[\frac{1}{5} + \frac{n}{25}]$ = 0 when $\frac{1}{5} + \frac{n}{25}$ < 1 ⇒ n < 20

$[\frac{1}{5} + \frac{n}{25}]$ = 1 when 1 ≤ $\frac{1}{5} + \frac{n}{25}$ < 2 ⇒ 20 ≤ n < 45

$\sum_{n = 1}^{44} [\frac{1}{5} + \frac{n}{25}]$ = $\sum_{n = 1}^{19} [\frac{1}{5} + \frac{n}{25}]$ + $\sum_{n = 20}^{44} [\frac{1}{5} + \frac{n}{25}]$ = 0 + (1 + 1 + 1 + … + 1 (25 times)) = 25

∴ N = 44

Hence, 44.

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