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
Approach Solution - 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 \(N = 44\).
Approach Solution -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:
\[ \boxed{N = 44} \]
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