To solve this problem, we need to express 55 as the sum of $n$ consecutive integers and then find the difference between the maximum and minimum values of $n$.

Step-by-Step Solution

1. Define Consecutive Integers: Suppose we have $n$ consecutive integers starting from some integer $a$. The sequence of these integers would be: $a, a+1, a+2, \ldots, a+(n-1)$

2. Sum of $n$ Consecutive Integers: The sum of $n$ consecutive integers starting from $a$ can be represented as: $S = a + (a + 1) + (a + 2) + \ldots + (a + n - 1)$ Using the formula for the sum of an arithmetic sequence, we have: $S = \frac{n}{2} \times (2a + n - 1)$ We are given that this sum $S = 55$. So, $\frac{n}{2} \times (2a + n - 1) = 55$

3. Solving for $n$ and $a$: Rearranging the equation: $n \times (2a + n - 1) = 110$ We need to find pairs of $n$ and $a$ such that this equation holds and results in consecutive integers.

4. Testing Possible Values of $n$: We’ll try different values of $n$ that divide 110 to see which satisfy the equation.

   • For $n = 10$: $2a + 10 - 1 = \frac{110}{10} = 11$ $2a = 11 - 9 = 2 \Rightarrow a = 1$ So, the integers are $1, 2, 3, \ldots, 10$, and the maximum and minimum are 10 and 1, respectively.

5. Calculate $\max(n) - \min(n)$: $\max(n) - \min(n) = 10 - 1 = 9$

Final Answer

The value of $\max(n) - \min(n)$ is 9.

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Related Questions

1. How do we determine if a number can be written as the sum of consecutive integers?
2. What other values of $n$ would work if the sum were a different number?
3. How can we generalize the approach to any number $S$ for finding consecutive integers?
4. Can we use this method to find sequences of consecutive integers for sums that aren't integers?
5. Why does the difference between $\max(n)$ and $\min(n)$ yield a consistent result in problems like these?

Tip

When dealing with consecutive integers summing to a particular number, break down the problem by dividing the sum by possible values of $n$, which helps in isolating the initial integer $a$.