Given the equation: \(x^2 - 2^y = 2023\)
Step 1. By trial, we find that \( x = 45 \) and \( y = 1 \) satisfy the equation, as:
\(45^2 - 2^1 = 2025 - 2 = 2023\)
Step 2. Thus, the only solution in \( C \) is \( (x, y) = (45, 1) \).
Step 3. Calculate \( \sum_{(x, y) \in C} (x + y) \):
\(\sum_{(x, y) \in C} (x + y) = 45 + 1 = 46\)
The Correct Answer is: 46
We are given the equation: \[ x^2 - 2^y = 2023 \] and are asked to find the sum \( \sum_{(x,y) \in C} (x + y) \).
The equation is: \[ x^2 - 2^y = 2023 \] Rearranging for \( x \) and \( y \), we get: \[ x^2 = 2023 + 2^y \] To find integer solutions for \( x \) and \( y \), we trial values for \( y \) to see which one yields a perfect square for \( x^2 \).
We start by testing small integer values for \( y \): - For \( y = 1 \), we get: \[ x^2 = 2023 + 2^1 = 2023 + 2 = 2025 \] \[ \sqrt{2025} = 45 \quad \Rightarrow \quad x = 45 \] Thus, the solution for \( x \) and \( y \) is \( x = 45 \) and \( y = 1 \).
The sum of \( x \) and \( y \) is: \[ x + y = 45 + 1 = 46 \]
The correct answer is: \[ \boxed{46} \]
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
