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>. | ||
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+ | Faster solution | ||
+ | Or you could also see that the different between 9 and 10 is 1, difference between 22 and 25 is 3, and add them up. | ||
==Query== | ==Query== |
Revision as of 08:32, 7 March 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?
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: .
Faster solution Or you could also see that the different between 9 and 10 is 1, difference between 22 and 25 is 3, and add them up.
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.