Let `f(x)=x^2-bx+c,b` is an odd positive integer. Given that f(x)=0 has two prime numbers as roots and b+c=35. If the least value of `f(x) AA x in R" is "lambda," then "[|lambda/3|]` is equal to (where [.] denotes greatest integer function)

Answer

To solve the problem step by step, we will follow the given information and derive the necessary values.

### Step 1: Understand the Function
We are given the function:
\[ f(x) = x^2 - bx + c \]
where \( b \) is an odd positive integer. The roots of this function are two prime numbers.

### Step 2: Use Vieta's Formulas
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = b \)
- The product of the roots \( \alpha \beta = c \)

### Step 3: Set Up the Equations
Given that \( b + c = 35 \), we can express \( c \) in terms of \( b \):
\[ c = 35 - b \]

### Step 4: Substitute \( c \) in Product of Roots
From Vieta's, we have:
\[ \alpha \beta = c = 35 - b \]

### Step 5: Identify Prime Roots
Let’s assume the two prime roots are \( p_1 \) and \( p_2 \). Thus:
\[ p_1 + p_2 = b \]
\[ p_1 \cdot p_2 = 35 - b \]

### Step 6: List Possible Prime Pairs
We need to find pairs of prime numbers whose sum is odd (since \( b \) is odd). The only way to get an odd sum from two primes is if one of them is 2 (the only even prime). Therefore, we can set:
- \( p_1 = 2 \)
- \( p_2 \) is an odd prime.

### Step 7: Find \( p_2 \)
Now, substituting \( p_1 = 2 \) into the equations:
\[ 2 + p_2 = b \]
\[ 2 \cdot p_2 = 35 - b \]

Substituting \( b \) from the first equation into the second:
\[ 2p_2 = 35 - (2 + p_2) \]
\[ 2p_2 = 33 - p_2 \]
\[ 3p_2 = 33 \]
\[ p_2 = 11 \]

### Step 8: Calculate \( b \) and \( c \)
Now substituting \( p_2 \) back to find \( b \):
\[ b = 2 + 11 = 13 \]
Then, substituting \( b \) into \( c = 35 - b \):
\[ c = 35 - 13 = 22 \]

### Step 9: Find the Minimum Value of \( f(x) \)
The function can be rewritten as:
\[ f(x) = x^2 - 13x + 22 \]
To find the minimum value of this quadratic function, we use the vertex formula:
\[ x = -\frac{b}{2a} = \frac{13}{2} \]

Now substituting \( x = \frac{13}{2} \) into \( f(x) \):
\[ f\left(\frac{13}{2}\right) = \left(\frac{13}{2}\right)^2 - 13 \cdot \frac{13}{2} + 22 \]
\[ = \frac{169}{4} - \frac{169}{2} + 22 \]
\[ = \frac{169}{4} - \frac{338}{4} + \frac{88}{4} \]
\[ = \frac{169 - 338 + 88}{4} = \frac{-81}{4} \]

### Step 10: Calculate \( \lambda \) and the Greatest Integer Function
Thus, the least value of \( f(x) \) is:
\[ \lambda = -\frac{81}{4} \]
Now, we need to find \( \left|\frac{\lambda}{3}\right| \):
\[ \left|\frac{-81/4}{3}\right| = \left|\frac{-81}{12}\right| = \frac{81}{12} = \frac{27}{4} = 6.75 \]

The greatest integer function \( [6.75] = 6 \).

### Final Answer
Thus, the final answer is:
\[ \boxed{6} \]