2011 AMC 10A Problem 4

Adding up each sum would take a while. Since we only want the difference, look for what the two sums have in common.

1. Write out the start and end of each sum:
- $X = 10 + 12 + 14 + \cdots + 98 + 100$
- $Y = \phantom{10 + {}}12 + 14 + \cdots + 98 + 100 + 102$
2. Both sums contain every even number from $12$ up to $100$.
3. The only differences: $X$ also has a $10$ at the start, and $Y$ also has a $102$ at the end.
4. When you subtract, all the shared numbers cancel. What is left is
$$Y - X = 102 - 10 = 92.$$

The answer is $\boxed{\textbf{(A)}\ 92}$.

When two sums are almost the same, don't add up either one. Line them up, cross out what they share, and subtract only what is different. The same "cancel the overlap" idea shows up in many harder sum problems.

Match the terms in order: $12$ with $10$, $14$ with $12$, and so on, up to $102$ with $100$. Each term of $Y$ is $2$ more than the matching term of $X$.

Count the terms of $X$: from $10$ to $100$ in steps of $2$ there are $\frac{100 - 10}{2} + 1 = 46$ numbers. So $Y$ is ahead by $2$ a total of $46$ times, and $Y - X = 46 \times 2 = 92$.

Both sums run through the even numbers $12, 14, \dots, 100$. Beyond that, $X$ has the extra term $10$ at the front and $Y$ has the extra term $102$ at the end.

When we subtract, every shared term cancels:
$$Y - X = 102 - 10 = 92.$$

The answer is $\boxed{\textbf{(A)}\ 92}$.