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

We need to find the value of N such that the sum equals 25.

The notation [x] means the floor function - the largest integer less than or equal to x. For example:

[2.7] = 2

[5.0] = 5

[0.9] = 0

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Let us first simplify what's inside the floor function:

$\tfrac{1}{5} + \tfrac{n}{25} = \tfrac{5}{25} + \tfrac{n}{25} = \tfrac{5+n}{25}$

So our sum becomes: $\sum_{n=1}^{N} \left[\tfrac{5+n}{25}\right] = 25$

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Now we need to see when $\left[\tfrac{5+n}{25}\right]$ changes value. Let us check different ranges:

For n = 1 to 19:

When n = 1: $\left[\tfrac{6}{25}\right] = [0.24] = 0$

When n = 19: $\left[\tfrac{24}{25}\right] = [0.96] = 0$

All terms from n = 1 to 19 contribute 0 to the sum.

For n = 20 to 44:

When n = 20: $\left[\tfrac{25}{25}\right] = [1.0] = 1$

When n = 44: $\left[\tfrac{49}{25}\right] = [1.96] = 1$

All terms from n = 20 to 44 contribute 1 to the sum.

For n = 45 onwards:

When n = 45: $\left[\tfrac{50}{25}\right] = [2.0] = 2$

Terms from n = 45 onwards contribute 2 or more to the sum.

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Since we want the total sum to be exactly 25:

Terms from n = 1 to 19: contribute 0 each = 0 total

Terms from n = 20 to 44: contribute 1 each = 25 total (since there are 25 terms: 44 - 20 + 1 = 25)

Perfect! The sum from n = 20 to 44 gives us exactly 25.

If we included n = 45, we would get 25 + 2 = 27, which exceeds our target.

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Therefore, N = 44.

The floor function creates "steps" in our sum. By understanding where these steps occur (at multiples of 25), we can quickly identify which terms contribute what values without tedious calculation.