Answer :
To find the greatest number of consecutive positive integers whose sum is 585, we can follow these steps:
1. Understand the problem:
- We need to express 585 as the sum of the maximum number of consecutive positive integers.
2. Formulate the sum of consecutive integers:
- Suppose we have \(k\) consecutive integers starting from \(x\).
- The integers would be \(x, x+1, x+2, \ldots, x+(k-1)\).
- The sum \(S\) of these \(k\) integers can be expressed as:
[tex]\[ S = x + (x+1) + (x+2) + \ldots + (x+(k-1)) \][/tex]
3. Simplify the sum:
- This sum can be written as:
[tex]\[ S = xk + (0 + 1 + 2 + \ldots + (k-1)) \][/tex]
- The sum of the first \(k\) integers is \(\frac{k(k-1)}{2}\).
- Hence,
[tex]\[ S = kx + \frac{k(k-1)}{2} \][/tex]
- Since \(S\) must equal 585,
[tex]\[ kx + \frac{k(k-1)}{2} = 585 \][/tex]
4. Solve for \(x\):
- Rearrange the equation to solve for \(x\):
[tex]\[ kx + \frac{k(k-1)}{2} = 585 \][/tex]
[tex]\[ kx = 585 - \frac{k(k-1)}{2} \][/tex]
[tex]\[ x = \frac{585 - \frac{k(k-1)}{2}}{k} \][/tex]
5. Determine feasibility:
- For \(x\) to be a positive integer, \(\frac{585 - \frac{k(k-1)}{2}}{k}\) must be a positive integer.
- We will need to check various values of \(k\) and see when \(x\) is a positive integer.
6. Check maximum value of \(k\):
- Upon examining \(k\) and ensuring that \(x\) remains a positive integer, we find that the maximum value for \(k\) that satisfies this condition is indeed 30.
Therefore, the greatest number of consecutive positive integers whose sum is 585 is:
Option (B) 30.
1. Understand the problem:
- We need to express 585 as the sum of the maximum number of consecutive positive integers.
2. Formulate the sum of consecutive integers:
- Suppose we have \(k\) consecutive integers starting from \(x\).
- The integers would be \(x, x+1, x+2, \ldots, x+(k-1)\).
- The sum \(S\) of these \(k\) integers can be expressed as:
[tex]\[ S = x + (x+1) + (x+2) + \ldots + (x+(k-1)) \][/tex]
3. Simplify the sum:
- This sum can be written as:
[tex]\[ S = xk + (0 + 1 + 2 + \ldots + (k-1)) \][/tex]
- The sum of the first \(k\) integers is \(\frac{k(k-1)}{2}\).
- Hence,
[tex]\[ S = kx + \frac{k(k-1)}{2} \][/tex]
- Since \(S\) must equal 585,
[tex]\[ kx + \frac{k(k-1)}{2} = 585 \][/tex]
4. Solve for \(x\):
- Rearrange the equation to solve for \(x\):
[tex]\[ kx + \frac{k(k-1)}{2} = 585 \][/tex]
[tex]\[ kx = 585 - \frac{k(k-1)}{2} \][/tex]
[tex]\[ x = \frac{585 - \frac{k(k-1)}{2}}{k} \][/tex]
5. Determine feasibility:
- For \(x\) to be a positive integer, \(\frac{585 - \frac{k(k-1)}{2}}{k}\) must be a positive integer.
- We will need to check various values of \(k\) and see when \(x\) is a positive integer.
6. Check maximum value of \(k\):
- Upon examining \(k\) and ensuring that \(x\) remains a positive integer, we find that the maximum value for \(k\) that satisfies this condition is indeed 30.
Therefore, the greatest number of consecutive positive integers whose sum is 585 is:
Option (B) 30.
