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 \( 64 \).
If \[ \sum_{r=1}^{30} r^2 \left( \binom{30}{r} \right)^2 = \alpha \times 2^{29}, \] then \( \alpha \) is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,