1. **Stating the problem:** We have two consecutive even positive numbers, and $L$ is their average. We want to compare $A=2$ with $B$, where $B$ is the remainder when $L$ is divided by 2. 2. **Define the consecutive even numbers:** Let the first even number be $2n$ where $n$ is a positive integer. The next consecutive even number is $2n+2$. 3. **Calculate the average $L$:** $$L=\frac{2n + (2n+2)}{2} = \frac{4n+2}{2} = 2n + 1$$ 4. **Analyze $L$:** Since $L = 2n + 1$, it is always an odd integer. 5. **Find the remainder $B$ when $L$ is divided by 2:** $$B = L \bmod 2 = (2n + 1) \bmod 2 = 1$$ 6. **Compare $A$ and $B$:** - $A = 2$ - $B = 1$ Since $2 > 1$, the answer is $A > B$. **Final answer:** $A$ is greater than $B$.