2008 AMC 12B Problems/Problem 2

The following problem is from both the 2008 AMC 12B #2 and 2008 AMC 10B #2, so both problems redirect to this page.

Problem

A $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?

$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

Solution 1

After reversing the numbers on the second and fourth rows, the block will look like this:

$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$

The positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{\textbf{(B) }}4$ .

Solution 2

You can also see that the positive difference between $9$ and $10$ is $1$ a d the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\boxed{\textbf{(B) }}4$

2008 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2008 AMC 12B Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

See Also