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 | + | The difference between the two diagonal sums is: <math>(4+9+16+25)-(1+10+17+22)=3-1-1+3=4 \Rightarrow A</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}} |
Revision as of 22:06, 15 May 2009
- 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 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?
Solution
After reversing the numbers on the second and fourth rows, the block will look like this:
The difference between the two diagonal sums is: .
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 |