Difference between revisions of "2008 AMC 12B Problems/Problem 2"
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The difference between the two diagonal sums is: <math>(4+9+16+25)-(1+10+17+22)=3-1-1+3=4 \Rightarrow B</math>. | The difference between the two diagonal sums is: <math>(4+9+16+25)-(1+10+17+22)=3-1-1+3=4 \Rightarrow B</math>. | ||
| + | |||
| + | ==Query== | ||
| + | If a different block of calendar dates had been shown, the answer would be unchanged. Why? | ||
==See Also== | ==See Also== | ||
Revision as of 23:55, 21 July 2020
- The following problem is from both the 2008 AMC 12B #2 and 2008 AMC 10B #2, so both problems redirect to this page.
Contents
Problem
A 
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?
![$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$](http://latex.artofproblemsolving.com/b/9/b/b9b73fd2d51a903cdaf987fca13b101dde70d13c.png)
Solution
After reversing the numbers on the second and fourth rows, the block will look like this:
![$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$](http://latex.artofproblemsolving.com/f/a/3/fa3f4afefd9e2d1f2e7e867c0c2df60feae30bef.png)
The difference between the two diagonal sums is: 
.
Query
If a different block of calendar dates had been shown, the answer would be unchanged. Why?
See Also
| 2008 AMC 12B (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 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. 
