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. 12
B. 24
C. 45
D. 54

Explanation

To solve the problem, we need to analyze 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 sum of the first \(n\) odd numbers is given by the formula:

\[ S_n = n^2 \]

### Step 2: Write the sums using the formula
Using the formula for the sum of the first \(n\) odd numbers, we can rewrite the equation as:

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

### Step 3: Recognize the equation
The equation \(p^2 + q^2 = r^2\) is a form of the Pythagorean theorem, which suggests that \(p\), \(q\), and \(r\) can be thought of as the lengths of the sides of a right triangle.

### Step 4: Find integer solutions
We are looking for integer values of \(p\), \(q\), and \(r\) such that \(p > 5\).

Let's start with the smallest integer greater than 5, which is \(p = 6\).

### Step 5: Check for Pythagorean triples
We need to find integers \(q\) and \(r\) such that:

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

This simplifies to:

\[ 36 + q^2 = r^2 \quad \Rightarrow \quad r^2 - q^2 = 36 \]

Factoring gives:

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

### Step 6: Find pairs \((r - q, r + q)\)
Now we can find pairs of factors of 36:

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

### Step 7: Solve for \(r\) and \(q\)
For each pair, we can solve for \(r\) and \(q\):

1. For \(1 \times 36\):
- \(r - q = 1\)
- \(r + q = 36\)
- Solving gives \(r = 18.5\) (not an integer)

2. For \(2 \times 18\):
- \(r - q = 2\)
- \(r + q = 18\)
- Solving gives \(r = 10\) and \(q = 8\)

3. For \(3 \times 12\):
- \(r - q = 3\)
- \(r + q = 12\)
- Solving gives \(r = 7.5\) (not an integer)

4. For \(4 \times 9\):
- \(r - q = 4\)
- \(r + q = 9\)
- Solving gives \(r = 6.5\) (not an integer)

5. For \(6 \times 6\):
- \(r - q = 6\)
- \(r + q = 6\)
- Solving gives \(r = 6\) and \(q = 0\) (not valid since \(q\) must be positive)

### Step 8: Valid solution
The only valid integers we found are \(p = 6\), \(q = 8\), and \(r = 10\).

### Step 9: Calculate \(p + q + r\)
Now we calculate:

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

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

\[ \boxed{24} \]