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

We are given the following sum: 

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

 First, let's find the expression inside the square brackets: \(\frac{1}{5} + \frac{n}{25} = \frac{5n + 1}{25}\) 

Now we want to find the largest integer less than or equal to \(\frac{5n + 1}{25}\), which is represented as \(\left[ \frac{5n + 1}{25} \right]\). 

We are given that for \(n = 1\) to \(n = 19\), the value of the function is zero. 

This means that \(\left[ \frac{5n + 1}{25} \right] = 0\) for \(n = 1\) to \(n = 19\). 

For \(n = 20\) to \(n = 44\), the value of the function is 1.  

This means that \(\left[ \frac{5n + 1}{25} \right] = 1\) for \(n = 20\) to \(n = 44\). 

Now, we want to find the value of \(N\) such that the sum of these bracketed terms is equal to 25. 

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

So, the value of \(N\) that satisfies the given equation is indeed \(N = 44\). 

To summarize: \(\sum_{n=1}^N \left[ \frac{1}{5} + \frac{n}{25} \right] = 25\) is satisfied when \(N = 44\).

Answer 2

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

Step 1: Simplify the expression inside the floor function

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

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

Step 2: Analyze when the floor value changes

• For \( n = 1 \) to \( n = 19 \):
  \( \frac{5 + n}{25} < \frac{24}{25} < 1 \)
  So, \( \left\lfloor \frac{5 + n}{25} \right\rfloor = 0 \)
• For \( n = 20 \) to \( n = 44 \):
  \( \frac{5 + 20}{25} = \frac{25}{25} = 1 \) and
  \( \frac{5 + 44}{25} = \frac{49}{25} = 1.96 \)
  So, the floor value is \( 1 \)

From \( n = 20 \) to \( n = 44 \), each term contributes 1 to the sum.

Step 3: Count the number of terms contributing 1

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

Thus, \[ \sum_{n=1}^{44} \left\lfloor \frac{5 + n}{25} \right\rfloor = 25 \]

Final Answer:

\[ \boxed{N = 44} \]