if `(1+3+5+7+....(2p-1))+(1+3+5+...+(2q-1)) =1+3+5+...+(2r -1),` then least possible value of `p+q+r `(Given `p>5`) is:

Options

A. 24
B. 32
C. 27
D. 25

Answer

To solve the problem step by step, we start with the given equation:

\[ (1 + 3 + 5 + \ldots + (2p - 1)) + (1 + 3 + 5 + \ldots + (2q - 1)) = 1 + 3 + 5 + \ldots + (2r - 1) \]

### Step 1: Understand the series
The series \(1 + 3 + 5 + \ldots + (2n - 1)\) represents the sum of the first \(n\) odd numbers. The formula for the sum of the first \(n\) odd numbers is:

\[ S_n = n^2 \]

### Step 2: Apply the formula to each series
Using the formula, we can express each part of the equation:

- The first series: \(1 + 3 + 5 + \ldots + (2p - 1) = p^2\)
- The second series: \(1 + 3 + 5 + \ldots + (2q - 1) = q^2\)
- The right-hand side series: \(1 + 3 + 5 + \ldots + (2r - 1) = r^2\)

### Step 3: Rewrite the equation
Substituting these sums into the original equation gives:

\[ p^2 + q^2 = r^2 \]

### Step 4: Rearranging the equation
We can rearrange this equation to find a relationship between \(p\), \(q\), and \(r\):

\[ p^2 + q^2 = r^2 \]

### Step 5: Identify the conditions
We are given the condition \(p > 5\). To minimize \(p + q + r\), we can start by assuming the smallest possible value for \(p\) that satisfies this condition, which is \(p = 6\).

### Step 6: Substitute \(p = 6\) into the equation
Substituting \(p = 6\) into the equation:

\[ 6^2 + q^2 = r^2 \]

This simplifies to:

\[ 36 + q^2 = r^2 \]

### Step 7: Rearranging to find \(r\)
Rearranging gives:

\[ r^2 - q^2 = 36 \]

This can be factored as:

\[ (r - q)(r + q) = 36 \]

### Step 8: Find pairs \((r - q)\) and \((r + q)\)
To solve for \(r\) and \(q\), we can consider the factor pairs of \(36\):

1. \(1 \times 36\)
2. \(2 \times 18\)
3. \(3 \times 12\)
4. \(4 \times 9\)
5. \(6 \times 6\)

For each factor pair \((a, b)\), we can set:

\[ r - q = a \quad \text{and} \quad r + q = b \]

From these, we can solve for \(r\) and \(q\):

\[ r = \frac{a + b}{2}, \quad q = \frac{b - a}{2} \]

### Step 9: Check valid pairs
We need \(q\) to be a positive integer. Let's check the pairs:

1. For \( (1, 36) \):
- \(r = \frac{1 + 36}{2} = 18.5\) (not valid)

2. For \( (2, 18) \):
- \(r = \frac{2 + 18}{2} = 10\)
- \(q = \frac{18 - 2}{2} = 8\) (valid)

3. For \( (3, 12) \):
- \(r = \frac{3 + 12}{2} = 7.5\) (not valid)

4. For \( (4, 9) \):
- \(r = \frac{4 + 9}{2} = 6.5\) (not valid)

5. For \( (6, 6) \):
- \(r = \frac{6 + 6}{2} = 6\)
- \(q = \frac{6 - 6}{2} = 0\) (not valid)

### Step 10: Calculate \(p + q + r\)
The only valid solution is \(p = 6\), \(q = 8\), and \(r = 10\).

Now we can find:

\[ p + q + r = 6 + 8 + 10 = 24 \]

### Final Answer
Thus, the least possible value of \(p + q + r\) is:

\[ \boxed{24} \]