Difference between revisions of "2013 AMC 8 Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
| − | We can simply test out numbers to see which one works. We can see that Clara’s score can’t be a multiple of ten because the reverse of the score is a one digit number, to small for the answer choices. After testing multiples, the answer should be <math>\textbf{ | + | We can simply test out numbers to see which one works. We can see that Clara’s score can’t be a multiple of ten because the reverse of the score is a one digit number, to small for the answer choices. After testing multiples, the answer should be <math>\textbf{A}</math> |
Note: Don’t use this method on a actual test unless you have a lot of time or just checking your work. | Note: Don’t use this method on a actual test unless you have a lot of time or just checking your work. | ||
Revision as of 20:49, 1 January 2023
Problem
When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?
Video Solution
https://www.youtube.com/watch?v=9FkjSCcdTqY
https://youtu.be/KBM2YN4kKGA ~savannahsolver
Solution
Let the two digits be 
and 
.
The correct score was 
. Clara misinterpreted it as 
. The difference between the two is 
which factors into 
. Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is 
.
Solution 2
We can simply test out numbers to see which one works. We can see that Clara’s score can’t be a multiple of ten because the reverse of the score is a one digit number, to small for the answer choices. After testing multiples, the answer should be 
Note: Don’t use this method on a actual test unless you have a lot of time or just checking your work.
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
