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2008 AMC 12B Problems/Problem 2

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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. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?

\begin{tabular}[t]{|c|c|c|c|}\multicolumn{4}{c}{}\\\hline1&2&3&4\\\hline8&9&10&11\\\hline15&16&am...

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

Solution

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}{}\\\hline1&2&3&4\\\hline11&10&9&8\\\hline15&16&am...

The difference between the two diagonal sums is: (4+9+16+25)-(1+10+17+22)=3-1-1+3=4 \Rightarrow B.

See Also

2008 AMC 12B (ProblemsResources)
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
2008 AMC 10B (ProblemsResources)
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
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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