Difference between revisions of "2008 AMC 12B Problems/Problem 2"
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==Problem== | ==Problem== | ||
| − | A <math>4\times 4</math> block of calendar dates is shown. | + | A <math>4\times 4</math> 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? |
<math>\begin{tabular}[t]{|c|c|c|c|} | <math>\begin{tabular}[t]{|c|c|c|c|} | ||
| Line 14: | Line 14: | ||
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math> | <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math> | ||
| − | ==Solution== | + | ==Solution 1== |
After reversing the numbers on the second and fourth rows, the block will look like this: | After reversing the numbers on the second and fourth rows, the block will look like this: | ||
| Line 25: | Line 25: | ||
\end{tabular}</math> | \end{tabular}</math> | ||
| − | The difference between the two diagonal sums is | + | The positive difference between the two diagonal sums is then <math>(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{\textbf{(B) } 4}</math>. |
| + | |||
| + | ==Solution 2== | ||
| + | Notice that at baseline the diagonals sum to the same number (<math>52</math>). Therefore we need only compute the effect of the swap. The positive difference between <math>9</math> and <math>10</math> is <math>1</math> and the positive difference between <math>22</math> and <math>25</math> is <math>3</math>. Adding gives <math>1+3=\boxed{\textbf{(B) } 4}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=1|num-a=3}} | {{AMC12 box|year=2008|ab=B|num-b=1|num-a=3}} | ||
{{AMC10 box|year=2008|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2008|ab=B|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:53, 31 July 2023
- 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. 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?
![$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$](http://latex.artofproblemsolving.com/b/9/b/b9b73fd2d51a903cdaf987fca13b101dde70d13c.png)
Solution 1
After reversing the numbers on the second and fourth rows, the block will look like this:
![$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$](http://latex.artofproblemsolving.com/f/a/3/fa3f4afefd9e2d1f2e7e867c0c2df60feae30bef.png)
The positive difference between the two diagonal sums is then 
.
Solution 2
Notice that at baseline the diagonals sum to the same number (
). Therefore we need only compute the effect of the swap. The positive difference between 
and 
is 
and the positive difference between 
and 
is 
. Adding gives 
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. 
