Question

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

Answer 1

Correct Answer: 44

We are given the following sum:

\[ \sum_{n=1}^N \left\lfloor \frac{1}{5} + \frac{n}{25} \right\rfloor = 25 \]

First, let's simplify the expression inside the brackets:

\[ \frac{1}{5} + \frac{n}{25} = \frac{5n + 1}{25} \]

Now, we need the greatest integer less than or equal to this value:

\[ \left\lfloor \frac{5n + 1}{25} \right\rfloor \]

We are told that for values of \( n = 1 \) to \( n = 19 \), the floor function evaluates to 0:

\[ \left\lfloor \frac{5n + 1}{25} \right\rfloor = 0 \quad \text{for } n = 1 \text{ to } 19 \]

For \( n = 20 \) to \( n = 44 \), the floor value becomes 1:

\[ \left\lfloor \frac{5n + 1}{25} \right\rfloor = 1 \quad \text{for } n = 20 \text{ to } 44 \]

We now compute the total sum:

\[ \sum_{n=1}^{N} \left\lfloor \frac{5n + 1}{25} \right\rfloor = \sum_{n=1}^{19} 0 + \sum_{n=20}^{44} 1 = 0 + 25 = 25 \]

So, the value of \( N \) that satisfies the equation is:

\[ \boxed{N = 44} \]

Answer 2

Given:

\[ \sum_{n=1}^{N} \left\lfloor \frac{1}{5} + \frac{n}{25} \right\rfloor = 25 \]

Step 1: Simplify the floor expression

We simplify inside the floor function: \[ \left\lfloor \frac{1}{5} + \frac{n}{25} \right\rfloor = \left\lfloor \frac{5 + n}{25} \right\rfloor \]

So the given sum becomes: \[ \sum_{n=1}^{N} \left\lfloor \frac{5 + n}{25} \right\rfloor = 25 \]

Step 2: Understand the floor value behavior

Let’s analyze how the floor function behaves for values of \(n\).

• For \( n = 1 \) to \( 19 \): \[ \left\lfloor \frac{5+n}{25} \right\rfloor < 1 \Rightarrow \text{Value is } 0 \]
• For \( n = 20 \) to \( 44 \): \[ 25 \leq 5 + n < 50 \Rightarrow \left\lfloor \frac{5+n}{25} \right\rfloor = 1 \]

So, the floor value is:

• 0 for 19 values
• 1 for the next set

Step 3: Find how many 1s are needed to make the sum 25

Since only values from \(n = 20\) to \(n = 44\) contribute 1 to the sum, we find the count:

\[ \text{Number of terms} = 44 - 20 + 1 = 25 \]

So the total sum is: \[ 25 \times 1 = 25 \]

Step 4: Final Answer

The value of \( N \) at which the sum reaches 25 is: \[ \boxed{N = 44} \]