Let \( m = 6a \) and \( n = 6b \), where \( a \) and \( b \) are co-prime numbers.
We are given that \( m \) and \( n \) are two-digit numbers.
Thus: \[ 10 \leq m \leq 99 \quad \text{and} \quad 10 \leq n \leq 99 \] So: \[ 10 \leq 6a \leq 99 \quad \Rightarrow \quad 2 \leq a \leq 16 \] and \[ 10 \leq 6b \leq 99 \quad \Rightarrow \quad 2 \leq b \leq 16 \]
Thus, \( a \) and \( b \) are integers, and the pairs \( (a, b) \) where \( \gcd(a, b) = 1 \) and \( a<b \) are the valid solutions.
Now, consider the valid values of \( a \) and \( b \), where both are between 2 and 16 and co-prime.
The valid pairs are as follows:
- \( a = 2, b = 3, 5, 7, 9, 11, 13, 15 \)
- \( a = 3, b = 4, 5, 7, 8, 10, 11, 13, 14, 16 \)
- \( a = 4, b = 5, 7, 9, 11, 13, 14, 16 \)
- \( a = 5, b = 6, 7, 8, 9, 11, 13, 14, 15 \)
- \( a = 6, b = 7, 9, 11, 13, 15 \)
- \( a = 7, b = 8, 9, 10, 11, 13, 14, 16 \)
- \( a = 8, b = 9, 11, 13, 15 \)
- \( a = 9, b = 10, 11, 13, 14, 16 \)
- \( a = 10, b = 11, 13, 15 \)
- \( a = 11, b = 12, 13, 14, 15 \)
- \( a = 12, b = 13, 14, 15, 16 \)
- \( a = 13, b = 14, 15, 16 \)
- \( a = 14, b = 15, 16 \)
- \( a = 15, b = 16 \)
Thus, there are 64 such ordered pairs.
Therefore, the correct answer is \( 64 \).
Step 1: Let \( m = 6a \) and \( n = 6b \), where \( a \) and \( b \) are co-prime numbers.
Step 2: We are given that:
\[ m < n \quad \Rightarrow \quad a < b \]
Step 3: Since \( m \) and \( n \) are two-digit numbers, we have:
\[ 10 \leq m \leq 99 \quad \text{and} \quad 10 \leq n \leq 99 \] This implies: \[ 2 \leq a \leq 16 \quad \text{and} \quad 2 \leq b \leq 16 \]
Step 4: Now, since \( a < b \) and \( a \) and \( b \) are co-prime, we consider the following pairs of \( a \) and \( b \) that satisfy these conditions:
For \( a = 2 \), \( b = 3, 5, 7, 9, 11, 13, 15 \)
For \( a = 3 \), \( b = 4, 5, 7, 8, 10, 11, 13, 14, 16 \)
For \( a = 4 \), \( b = 5, 7, 9, 11, 13, 15 \)
For \( a = 5 \), \( b = 6, 7, 8, 9, 11, 12, 13, 15, 16 \)
For \( a = 6 \), \( b = 7, 11, 13 \)
For \( a = 7 \), \( b = 8, 9, 10, 11, 12, 13, 15, 16 \)
For \( a = 8 \), \( b = 9, 11, 13, 15 \)
For \( a = 9 \), \( b = 10, 11, 13, 14, 16 \)
For \( a = 10 \), \( b = 11, 13, 14, 16 \)
For \( a = 11 \), \( b = 12, 13, 14, 15, 16 \)
For \( a = 12 \), \( b = 13, 14, 15, 16 \)
For \( a = 13 \), \( b = 14, 15, 16 \)
For \( a = 14 \), \( b = 15, 16 \)
For \( a = 15 \), \( b = 16 \)
Step 5: Total number of ordered pairs is \( 64 \).
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: