2008 AMC 10B Problem 2

This is a bookkeeping problem, so the safest plan is to actually write the new block down instead of keeping it in your head.

1. Reverse the second row: $8, 9, 10, 11$ becomes $11, 10, 9, 8$.
2. Reverse the fourth row: $22, 23, 24, 25$ becomes $25, 24, 23, 22$.
3. The new block is
$$\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 11 & 10 & 9 & 8 \\ 15 & 16 & 17 & 18 \\ 25 & 24 & 23 & 22 \end{array}$$
4. Diagonal from top-left to bottom-right: $1 + 10 + 17 + 22 = 50$.
5. Diagonal from top-right to bottom-left: $4 + 9 + 16 + 25 = 54$.
6. The difference is $54 - 50 = 4$.

The answer is $\boxed{\textbf{(B)}\ 4}$.

Nothing clever is needed; the only danger is a copying slip, and writing the new block down removes it. You can double-check with a shortcut. In the original block both diagonals add to $52$ ($1 + 9 + 17 + 25$ and $4 + 10 + 16 + 22$). After the reversals, the top-left diagonal gets $10$ instead of $9$ (up $1$) and $22$ instead of $25$ (down $3$), so it drops by $2$ to $50$. The other diagonal rises by $2$ to $54$.

Reverse the second row ($8, 9, 10, 11 \to 11, 10, 9, 8$) and the fourth row ($22, 23, 24, 25 \to 25, 24, 23, 22$). The block becomes
$$\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 11 & 10 & 9 & 8 \\ 15 & 16 & 17 & 18 \\ 25 & 24 & 23 & 22 \end{array}$$

Main diagonal (top-left to bottom-right): $1 + 10 + 17 + 22 = 50$.

Other diagonal (top-right to bottom-left): $4 + 9 + 16 + 25 = 54$.

The positive difference is $54 - 50 = 4$.

The answer is $\boxed{\textbf{(B)}\ 4}$.