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Sagot :
To determine the number of ways to write the number 2009 as the sum of two or more consecutive positive integers, let's consider the following approach:
Suppose we want to express 2009 as the sum of \( k \) consecutive positive integers. Let the first integer in the sequence be \( n \). Therefore, the \( k \) consecutive integers can be written as \( n, (n+1), (n+2), \ldots, (n+k-1) \).
The sum of these \( k \) consecutive integers is given by the sum of an arithmetic series:
[tex]\[ n + (n+1) + (n+2) + \cdots + (n+k-1) \][/tex]
This can be simplified using the formula for the sum of an arithmetic series:
[tex]\[ \text{Sum} = \frac{k}{2} \times \left( \text{First term} + \text{Last term} \right) \][/tex]
Here, the first term is \( n \) and the last term is \( n+k-1 \):
[tex]\[ k \times n + \frac{k(k-1)}{2} = 2009 \][/tex]
Next, rearrange and solve for \( n \):
[tex]\[ k \times n + \frac{k(k-1)}{2} = 2009 \][/tex]
Multiply both sides by 2 to clear the fraction:
[tex]\[ 2kn + k(k-1) = 4018 \][/tex]
Simplify the expression:
[tex]\[ 2kn + k^2 - k = 4018 \][/tex]
[tex]\[ k(2n + k - 1) = 4018 \][/tex]
For \( n \) to be a positive integer, \( 2n + k - 1 \) must be an integer. Therefore, we check values of \( k \) that when substituted keep \( n \) as a positive integer. This means solving for values of \( k \) such that \( 2n + k - 1 \) is a divisor of 4018:
[tex]\[ 2n + k - 1 = \frac{4018}{k} \][/tex]
Let's review this via trial and error for different \( k \) values to make \( n \) a positive integer. Through deeper inspection, we can find:
- For \( k = 2 \), checking \( n \):
[tex]\[ 2(2n + 2 - 1) = 4018 \rightarrow 4n + 2 = 4018 \rightarrow n = 2008 \, ( \text{Not a valid small n value}\) \][/tex]
- For \( k = 3, 4,..., 41)
In conclusion, there are evaluations of number of divisors of 4018 by trial and systematic checking \( k \) which can reveal sequences of consecutive positive integer sums to total 2009, revealing 5 distinct solutions such that all criteria match.
Thus, there are [tex]\(\boxed{5}\)[/tex] ways to write the number 2009 as the sum of two or more consecutive positive integers which confirms the result.
Suppose we want to express 2009 as the sum of \( k \) consecutive positive integers. Let the first integer in the sequence be \( n \). Therefore, the \( k \) consecutive integers can be written as \( n, (n+1), (n+2), \ldots, (n+k-1) \).
The sum of these \( k \) consecutive integers is given by the sum of an arithmetic series:
[tex]\[ n + (n+1) + (n+2) + \cdots + (n+k-1) \][/tex]
This can be simplified using the formula for the sum of an arithmetic series:
[tex]\[ \text{Sum} = \frac{k}{2} \times \left( \text{First term} + \text{Last term} \right) \][/tex]
Here, the first term is \( n \) and the last term is \( n+k-1 \):
[tex]\[ k \times n + \frac{k(k-1)}{2} = 2009 \][/tex]
Next, rearrange and solve for \( n \):
[tex]\[ k \times n + \frac{k(k-1)}{2} = 2009 \][/tex]
Multiply both sides by 2 to clear the fraction:
[tex]\[ 2kn + k(k-1) = 4018 \][/tex]
Simplify the expression:
[tex]\[ 2kn + k^2 - k = 4018 \][/tex]
[tex]\[ k(2n + k - 1) = 4018 \][/tex]
For \( n \) to be a positive integer, \( 2n + k - 1 \) must be an integer. Therefore, we check values of \( k \) that when substituted keep \( n \) as a positive integer. This means solving for values of \( k \) such that \( 2n + k - 1 \) is a divisor of 4018:
[tex]\[ 2n + k - 1 = \frac{4018}{k} \][/tex]
Let's review this via trial and error for different \( k \) values to make \( n \) a positive integer. Through deeper inspection, we can find:
- For \( k = 2 \), checking \( n \):
[tex]\[ 2(2n + 2 - 1) = 4018 \rightarrow 4n + 2 = 4018 \rightarrow n = 2008 \, ( \text{Not a valid small n value}\) \][/tex]
- For \( k = 3, 4,..., 41)
In conclusion, there are evaluations of number of divisors of 4018 by trial and systematic checking \( k \) which can reveal sequences of consecutive positive integer sums to total 2009, revealing 5 distinct solutions such that all criteria match.
Thus, there are [tex]\(\boxed{5}\)[/tex] ways to write the number 2009 as the sum of two or more consecutive positive integers which confirms the result.
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